1. ELECTROSTATICS
The branch of physics which deals with charges at rest is called electrostatics.
2. ELECTRIC CHARGE
Charge is scalar physical quantity associated with matter due to which it produces and experiences electrical and magnetic effects. The excess or deficiency of electrons in a body gives the concept of charge. A negatively charged body has excess of electrons while a positively charged body has lost some of its electrons.
+ + + + + + + + + + + + + + + + + + + + ++ + + + + + -+ -+ -+ -+ -+ -+ + + + + + +
-Electrons = Protons Electrons < Protons Electrons > Protons Positively charged body Negatively charged body Properties of Charge
(1) Like charges repel while unlike charges attract each other. The true test of electrification is repulsion (2) Charge is a scalar.
(3) Charge is always associated with mass :
In charging, the mass of a body changes. If electrons are removed from the body, the mass of the body will decrease and the body will become positively charged. If electrons are added to a body, the mass of the body will increase and the body will acquire a net negative charge. Due to extremely small mass of electron (= 9.1 × 10–31 kg) the change in mass of a body due to charging is negligible as compared to the
mass of the body.
M + m > M Body = MNeutral + + + + + ++ + + + + + + + + + + M - m < M
ELECTRIC CHARGE AND FIELD
(4) Charge is quantised : When a physical quantity has only discrete values the quantity is said to be quantised. Milikan oil drop experiment established that the smallest charge that can exist in nature is the charge of an electron. If the charge of an electron (e = 1.6 × 10–19 C) is taken as the elementary unit, i.e., quanta
of charge, the charge on a body will be an integral multiple of e i.e., q = ± ne with n = 1, 2, ...
(5) Charge is invariant : This means that charge is independent of frame of reference, i.e., charge on a body does not change with speed. The charge density or mass of a body depends on speed and increases with increase in speed.
(6) Unit of Charge :
[a] S.I. : coulomb. [1 Coulomb = 1 ampere × 1 second] [b] C.G.S. : Static coulomb or frankline
1 coulomb = 3 × 109 static coulomb
1 coulomb = 3 × 109 esu of charge =
10 1
emu of charge [esu = electrostatic unit] [emu = electro magnetic unit]
Practical units of charge are amp × hr (= 3600 coulomb) and faraday (= 96500 coulomb)
3. CHARGING OF A BODY
(a) Friction : In friction when two bodies are rubbed together, electrons are transferred from one body to the other. This makes one body become positively charged while the other become negatively charged, e.g., when a glass rod is rubbed with silk, the rod becomes positively charged while the silk is negatively charged. Clouds are also charged by friction. Charging by friction is in accordance with conservation of charge. The positive and negative charges appear simultaneously in equal amounts due to transfer of electrons from one body to the other. (b) Induction : If a charged body is brought near a neutral body, the charged body will attract opposite charge
and repel similar charge present in the neutral body. This makes one side of the neutral body become positively charged while the other side negative.
Cha rgin g body ++ + + + + ++ +++ + q'= 0 V' = +iv e Cha rged body is brou ght n ear unch arge dbo dy + + + + + + + q' =-ive V' =0 Uncharg ed body is
connected to the Ea rth + + + + + + q' = -ive V' =0 Uncharge d body is disconnec ted fromthe earth q' = -ive V' = -ive Chargin g body is re moved Charging a body by induction
e
Important Points
(i) Inducing body neither gains nor loses charge.
(ii) The nature of induced charge is always opposite to that of inducing charge.
(iii) Induced charge can be lesser or equal to inducing charge (but never greater) and its maximum value is
K 1 1 q ' q
where q is the inducing charge and K is the dielectric constant of the material of the uncharged body. (iv) For metals, K = and so q' = –q i.e., in metals induced charge is equal and opposite to inducing charge. (v) Induction takes place only in bodies (either conducting or non conducting) and not in particles.
(c) Conduction : When an insulated conductor is brought in contact with a charged body and it gets the same charge as the charged body then conduction takes place. Conduction is only possible in conductors and not in insulators.
4. COULOMB'S LAW
The force of attraction or repulsion between two stationary point charges is directly proportional to the product of charges and inversely proportional to the square of distance between them. This force acts along the line joining the centre of two charges.
If q1 & q2 are charges, r is the distance between them and F is the force acting between them Then, F q1 q2, F 1/r² F 122 r q q or 2 2 1 r q q C F q1 q2 r
C is const. which depends upon system of units and also on medium between two charges ² C / ² Nm 10 9 4 1 C 9 0 (In SI unit)
C = 1 in electrostatic unit (esu)
0 = 8.85 × 10–12 C²/Nm² = permittivity of free space or vacuum
Effect of medium
The dielectric constant of a medium is the ratio of the electrostatic force between two charges separated by a given distance in air to electrostatic force between same two charges separated by same distance in that medium. Fair= 122 0 r q q 4 1 and Fmedium = 2 2 1 r 0 r q q 4 1 r air medium 1 F F = K
r or K = Dielectric constant or Relative permittivity or specific inductive capacity of medium.
(i) Permittivity : Permittivity is a measure of the ability of the medium surrounding electric charges to allow electric lines of force to pass through it. It determines the forces between the charges.
(ii) Relative Permittivity : The relative permittivity or the dielectric constant (r or K) of a medium is defined as the ratio of the permittivity of the medium to the permittivity 0 of free space i.e. r or
0 K Dimensions of permittivity 2 2 0 length F Q 2 222 L MLT A T = M–1 L–3 T4A2
The dielectric constants of different mediums are
Medium Vacuum Air Water Mica Teflon Glass PVC Metal
r 1 1.00059 80 6 2 5-10 4.5
Coulomb's law in vector form
The direction of the force acting between two charges depends on their nature and it is along the line joining the centre of two charges.
21 F = force on q2 due to q1 12 2 12 2 1 r 0 21 rˆ r q q 4 1 F 12 F = Force on q1 due to q2 21 2 21 2 1 r 0 12 rˆ r q q 4 1 F r12 F21 F12 q1 q2 21 12 F F (as rˆ12rˆ21) or F F 0 21 12 5. ELECTRIC FIELD
To explain 'action at a distance', i.e., 'force without contact' between charges we assume that a charge or charge distribution produces a field in space surrounding it. The region surrounding a charge or charge distribution in which its electrical effects are perceptable is called the electric field of the given charge. Electric field at a point is characterised either by a vector function of position
E called electric intensity or by a scalar function of position V called electric potential. The electric field in a certain space is also visualised graphically in terms of lines of force. So electric intensity, potential and lines of force are different ways of describing the same field
Electric field Intensity
E
The electric field intensity at a point in an electric field is defined as the force experienced by a unit positive point charge called test charge supposed to be placed at that point. The test charge does not affect the source charge or charge distribution producing the field. If a test charge q0 at a point P in an electric field experiences a force
F, then electric field ) q / F ( E 0
If the field is produced by a point charge q, then from Coulomb's law
r r qq 4 1 F 30 0
field due to point-charge
q at position r in free space 2 0 3 0 0 r q 4 1 E or r r q 4 1 q F E
If field is produced by a charge distribution, then by 'principle of superposition' field is given as Source Charge q O r P q0 q1 r1 P q2 q4 q2
n 1 i i 2 1 E ... E E E with 3 i i i 0 i r r q 4 1 Ewhile for continuous charge distribution (treating small charge element as a point charge), , r r dq 4 1 E d 3 0 i.e.,
r r dq 4 1 E 3 0 Important points(1) It is a vector quantity having dimensions 3 1
2 A MLT AT MLT q F E The SI unit is N/C or V/m as m V m C J m C m N C N Nm J J/CV (2) By definition E F/q0, or Fq E 0
A charged particle in an electric field experiences a force whether it is at rest or in motion. The direction of force is along the field if it is positive and opposite to the field if it is negative.
+ F = +qE
E
F = -qE E
(3) In free space Electric field is 2 0 0 r q 4 1 E
In a medium of permittivityfield is
2 r q 4 1 E So, K 1 E E 0 0 [as = 0 K] or, E = E0/K
In presence of a dielectric, electric field decreases and becomes 1/K times of its value in free space.
6. ELECTRIC LINES OF FORCE
The idea of lines of force was introduced by Michel Faraday. A line of force is an imaginary curve the tangent to which at a point gives the direction of intensity at that point and the number of lines of force per unit area normal to the surface surrounding that point gives the magnitude of intensity at that point.
A
EB
B
Important points
(1) Electric lines of force usually start or diverge out from positive charge and end or converge on negative charge.
+ –
(2) The number of lines originating or terminating on a charge is proportional to the magnitude of charge. In SI units 1/0 shows electric lines associated with unit (i.e., 1 coulomb) charge. So if a body encloses a charge q, total lines of force or flux associated with it is q/0. If the body is cubical and charge is situated at its centre the lines of force through each face will be q/60.
+q
Total lines of force (q/ ) while through each face (q/6 )
0 0
Charge A is positive while B is negative and q > qA B
A B
(3) Lines of force never cross each other because if they cross then intensity at that point will have two directions which is not possible.
(4) In electrostatics the electric lines of force can never be closed loops, as a line can never start and end on the same charge. If a line of force is a closed curve, work done round a closed path will not be zero and electric field will not remain conservative.
(5) Lines of force have tendency to contract longitudinally like a stretched elastic string producing attraction between opposite charges and repel each other laterally resulting in, repulsion between similar charges and 'edge-effect' (curving of lines of force near the edges of a charged conductor)
Electric lines of force for a system of two positive charges
–
Attraction
Electric lines of force for a dipole +
Repulsion
(6) If the lines of force are equidistant straight lines the field is uniform and if lines of force are not equidistant or straight lines or both, the field will be non-uniform. The first three represent non-uniform field while last shows uniform field.
Magnitude is not constant
Direction is not constant
Both magnitude and direction not constant
Both magnitude and direction constant (7) Electric lines of force end or start normally on the surface of a conductor. If a line of force is not
normal to the surface of a conductor, electric intensity will have a component along the surface of the conductor and hence conductor will not remain equipotential which is not possible as in electrostatics conductor is an equipotential surface.
Fixed point charge near infinite metal plate
(A)
+
d
+q
Parallel metal plates having dissimilar charges
(B) + + – + + + + + – – – – – – E = 0 E = 0 Edge Effect E = 0 Uniform field
Parallel metal plates having similar charges
+ + + + + + + E = 0 + + + + + + + (C) E = 0 E = 0
(8) If in a region of space, there is no electric field there will be no lines of force. This is why inside a conductor or at a neutral point where resultant intensity is zero there is no line of force.
(9) The number of lines of force per unit normal area at a point represents magnitude of electric field intensity. The crowded lines represent strong field while distant lines shows a weak field.
(10) The tangent to the line of force at a point in an electric field gives the direction of intensity. It gives direction of force and hence acceleration which a positive charge will experience there (and not the direction of motion). A positive point charge free to move may or may not follow the line of force. It will follow the line of force if it is a straight line (as direction of velocity and acceleration will be same) and will not follow the line if it is curved as the direction of motion will be different from that of acceleration. The particle will not move in the direction of motion or acceleration (line of force) but other than these which will vary with time as v uat.
7. ELECTRIC-FLUX
Electric flux through an elementary area ds is defined as the scalar product of area and field, i.e.,
E•ds Edscos
d E i.e., E
E•dsIt represents the total lines of force passing through the given area. Here area is treated as a vector. The direction of area vector is given by direction of normal to the surface.
Important points
(1) It is a real scalar physical quantity with units volt × m and dimensions
1 3 3 2 2 E L MLT A AT MLT ds q F Eds
(2) It will be maximum when cos is max = 1, i.e., = 0°, i.e., electric field is normal to the area with (dE)max = E ds
(3) It will be minimum when cos is min = 0, i.e., = 90°, i.e. field is parallel to the area with (dE)min = 0 (4) For a closed body outward flux is taken as positive while inward flux is taken as negative.
n E Positive - flux Body Negative - flux E n E n n
Cylinder in a uniform field
E R2 E 0 E E R2E Body E n^
8. GAUSS'S LAW
It relates the total flux of an electric field through a closed surface to the net charge enclosed by that surface. According to it, the total flux linked with a closed surface is 1/0 times the charge enclosed by the closed surface,
Mathematically 0 s q ds • E
Applications of Gauss law
(1) Electric field due to a line charge : Gauss law is useful in calculating electric field intensity due to symmetrical charge distributions.
We consider a gaussian surface which is a cylinder of radius r which encloses a line charge of length h with line charge density .
According to Gauss law
E.ds =0 in q surface l Cylindrica ds . E
+ surface circular ds . E I
+ surface circular ds . E II
= 0 h suface l Cylindrica º 0 cos Eds
+ surface circular cos Eds I
2 + surface circular cos Eds II
2 = 0 h E(2 r h) = 0 h So E = 2 r 0 ) r 1 E ( (2) Electric field due to an infinite plane thin sheet of charge :
To find electric field due to the plane sheet of charge at any point P distant r from it, choose a cylinder of area of cross-section A through the point P as the Gaussian surface. The flux due to the electric field of the plane sheet of charge passes only through the two circular caps of the cylinder. Let surface charge density =
According to gauss law
E.dS qin/00 surface l cylindrica surface circular II surface circular I A cos ds E cos ds E cos ds E
or EA + EA + 0 = 0 2 A or E = 0 2 r E +++ + + + +++ + + + + + + E P E Q r Plane sheet of charge Gaussian Surface(3) Electric field intensity due to uniformly charged spherical shell : We consider a thin shell of radius R carrying a charge Q on its surface (i) at a point P0outside the shell (r > R)
According to gauss law
1 S 0.ds E = 0 Q or E0 (4r 2 ) = 0 Q E0 = 2 0r 4 Q = 0 2 2 r Rwhere the surface charge density = surfacetotalchargareae = 2 R 4
Q
The electric field at any point outside the shell is same as if the entire charge is concentrated at centre of shell.
(ii) at a point Ps on surface of shell (r = R)
ES = 2 0R 4 Q = 0
(iii) at a point Pin inside the shell (r < R)
According to gauss law
2 S ds . E = 0 in q As enclosed charge qin= 0 So Ein= 0The electric field inside the spherical shell is always zero.
(4) Electric field intensity due to a spherical uniformly charge distribution :
We consider a spherical uniformly charge distribution of radius R in which total charge Q is uniformly distributed throughout the volume.
The charge density =
volume total e arg ch total = 3 R 3 4 Q = 4 R3 Q 3
(i) at a point P0 outside the sphere (r > R)
according to gauss law
E0.ds =0 Q or E0 (4r 2 ) = 0 Q or E0 = 2 0r 4 Q = 30 2 3 r R E = 0 E = 0
distance from centre (r)
Emax = Q/4 0R 2 E O r < R r = R r > R 2 r 1 E 2 r 1 E ds + R O Pin PSP0 + + + + + + + + + + + + + + + + + + + + + r
(ii) at a point Ps on surface of sphere (r = R) Es= 2 0R 4 Q = 30 R
(iii)at a point Pin inside the sphere (r < R) According to gauss law
Ein.ds = 0 in q = 0 1 .3 4 r3 = 3 0 3 R Qr Ein(4r2) = 3 0 3 R Qr or Ein = 3 0R 4 Qr = 30 r (Ein r) SOLVED EXAMPLES Ex. How many electrons are present in 1 coulomb charge.Sol. q = ne q = 1C e = 1.6 × 10–19 C So n = q/e = 6.25 × 1018 electrons.
Ex. A copper sphere of mass 2.0 g contains about 2 × 1022 atoms. The charge on the nucleus of each atom is 29e.
(i) How many electrons must be removed from the sphere to give it a charge of +2µC? (ii) Determine the fraction of electrons removed.
(iii) Is there any change in mass of sphere when it is given positive charge?
Sol. (i) Number of electrons to be removed 19 6 10 6 . 1 10 2 e Q n = 1.25 × 1013
(ii) Total number of electrons in the sphere = 29 × 2 × 1022 = 5.8 × 1023
Fraction of electrons removed 23 11
13 10 16 . 2 10 8 . 5 10 25 . 1
Thus 2.16 × 10–9 % of electrons are to be removed to give the sphere a charge of 2µC.
(iii) Yes mass decreases, when body is given a positive charge.
Decrease of mass m = 9 × 10–31 × 1.25 × 1013 = 1.125 × 10–17 kg
Ex. Consider four equal charges placed on the corners of a square with side a. Determine the magnitude and direction of the resultant force on the charge on lower right corner.
O r < R r = R r > R E 2 r 1 E r2 1 E E r E r
Sol. The forces on the charge on lower right corner due to charges 1, 2, 3 are F1 = kq²/a², F2 = kq²/a², F3 = kq²/2a². The resultant of F1 and F2 is
F F 2FF cos90
F12 12 22 1 2 = 2kq2/a2.
This is in the direction parallel to F3. Therefore the total force on the said charge is F = F12 + F3
1 2 2
a kq 2 1 F 2 2 The direction of F is, 45° below the horizontal line.
Ex. Three identical spheres each having a charge q and radius R, are kept in such a way that each touches the other two. Find the magnitude of the electric force on any sphere due to other two.
Sol. For external points a charged sphere behaves as if the whole of its charge was concentrated at its centre.
Force on A due to B is
alongBA R 4 q 4 1 R 2 q q 4 1 F 2 2 0 2 0 AB Force on A due to C.
alongCA R 4 q 4 1 R 2 q q 4 1 F 2 2 0 2 0 ACNow, angle between BA and CA is 60° and FAB = FAC = F.
F 3 60 cos FF 2 F F FA 2 2
2 0 R q 4 3 4 1 Ex. Two identical charged spheres are suspended by strings of equal length. The strings make an angle of 30° with each other. When suspended in a liquid of density 0.8 gm/cc, the angle remains the same. What is the dielectric constant of the liquid? ( = 1.6 gm/cc is the density of the sphere)
Sol. The forces acting on each ball are tension T, weight mg and electric force F, for its equilibrium along vertical, T cos = mg
and along horizontal T sin = F Dividing we have
mg F
tan ....(1)
When the balls are suspended in a liquid of density and dielectric constant K, the electric force will become (1/K) times, i.e., F' = (F/K) while weight mg' = mg – Th = mg – Vg [as Th = Vg]
i.e. mg 1 ' mg m V as
So for equilibrium of ball,
/ 1 Kmg F ' mg ' F ' tan ....(2)
According to given problem ' = ; so from eqn. (1) & (2), we have
2 8 . 0 6 . 1 6 . 1 K Ex. An infinite plane of positive charge has a surface charge density . A metal ball B of mass m and charge q is attached to a thread and tied to a point A on the sheet PQ. Find the angle which AB makes with the plane PQ.
Sol. Due to positive charge the ball will experience electrical force Fe = qE horizontally away from the sheet while the weight of the ball will act vertically downwards and hence if T is the tension in the string, for equilibrium of ball:
Along horizontal, T sin = qE And along vertical, T cos = mg So tan =
mg qE
and T = [(mg)2 + (qE)2]1/2
The field E produced by the sheet of charge PQ having charge density is
0 2 E So, mg 2 q tan 0 i.e., mg 2 q tan 0 1
Ex. A point charge q is placed at one corner of a cube of edge a. What is the flux through each face of the cube?
Sol. At a corner, 8 cubes can be placed symmetrically, flux linked with each cube due to a charge q at the corner will be q/80
For the faces passing through the edge A, electric field E will be parallel to area of face and so flux through these three faces will be zero. As the cube has six faces and flux linked with three faces (through A) is zero, so flux linked with remaining three face will be (q/80). The remaining
three faces are symmetrical so flux linked with each of the three faces passing through B will be,
0 0 q 24 1 q 8 1 3 1 .
Ex. Flux entering a closed surface is 2000 V-m. Flux leaving that surface is 8000 V-m. Find the charge inside surface. Sol. Net flux = out – in = (8000 – 2000) = 6000 V-m
0 q so q = (6000) (8.85 × 10–12) = 0.53 µC T A+ P Q mg qE q B + + + + + Z X Y L A q B
1. When two bodies A and B are rubbed with each other, A gets a positive charge. In this process – (1) Protons are transferred from A to B
(2) Protons are transferred from B to A (3) Electrons are transferred from A to B (4) Electrons are transferred from B to A 2. Choose the correct option from following –
(1) If a particle possesses mass, then it must posses charge
(2) If a particle possesses charge, then it must posses mass
(3) If a particle possesses mass, then it must be uncharged
(4) Both charge and mass for a particle cannot be zero
3. Five balls, numbered 1 to 5, are suspended using separated threads. Pairs (1, 2), (2, 4), (4, 1) show electrostatic attraction, while pairs (2, 3) and (4, 5) show repulsion, therefore ball 1–
(1) must be positively charged (2) must be negatively charged (3) may be neutral
(4) must be made of metal
4. An isolated solid metallic sphere is given +Q charge. The charge will be distributed on the sphere–
(1) uniformly but only on surface (2) only on surface but non uniformly (3) uniformly inside the volume (4) non uniformly inside the volume
5. A soap bubble is given a negative charge then its radius–
(1) decreases (2) increases
(3) remains unchanged
(4) nothing can be predicted as information is insufficient
EXERCISE – 1(A)
6. A sure test of electrification is– (1) attraction (2) repulsion (3) friction (4) induction 7. Mark correct option or options–
(1) like charged bodies always repel each other (2) like charged bodies always attract each other (3) like charged bodies may attract each other (4) none of the above
8. Dimensions of
mg
ke2
is same as that of (here k = Coulomb’s constant), e is charge of electron, m is mass of proton and g is acceleration due to gravity –
(1) Area (2) Pressure
(3) Energy (4) Volume
9. Two small balls with like charges are suspended by light strings of equal length L from the same point. When taken to a place where they are in a state of weightlessness the separation between the balls will
be-(1) 2L (2) L
2 (3)
L L
(
1
)
2
(4)L L
(
1
)
10. A, B, C are three identical small metal balls having
charges q, –3q and q respectively. When A and C are placed at a certain distance apart electrostatic force between them is F. If B is touched with A and then removed, then magnitude of electrostatic force between A and C will be
(1) 4 F (2) F (3) 2 F (4) 2F
11. Three charge +4q, Q and q are placed in a straight line of length l at points distance 0, l/2 and l respectively. What should be the value of Q in order to make the net force on q to be zero ?
(1) –q (2) –2q
12. Two similar very small conducting spheres having charges 40C and –20C are some distance apart. Now they are touched and kept at same distance. The ratio of the initial to the final force between them is–
(1) 8 : 1 (2) 4 : 1
(3) 1 : 8 (4) 1 : 1
13. Charges +Q, +Q and –Q are placed on the vertices A, B, C of a triangle ABC respectively. The side of equilateral triangle is a. Magnitude of force acting on charge at A is – (1) 2 2 a kQ (2) 3 2 2 a kQ (3) Zero (4) 2 2 2 a kQ
14. Two small spheres, each of mass 0.1 gm and carrying same charge 10–9 C are suspended by threads of equal length from the same point. If the distance between the centres of the sphere is 3 cm. Then find out the angle made by the thread with the vertical. (1) 50 1 tan 1 (2) 100 1 tan 1 (3) 150 1 tan 1 (4) 200 1 tan 1
15. Two identical conducting spheres (of negligible radius), having charges of opposite sign, attract each other with a force of 0.108 N when separated by 0.5 meter. The spheres are connected by a conducting wire, which is then removed (when charge stops flowing), and thereafter repel each other with a fore of 0.036 N keeping the distance same. What were the initial charges on the spheres?
(1) ± 1 × 10–6 C, ± 1 × 10–6 C (2) ± 1 × 10–6 C, ± 3 × 10–6 C (3) ± 3 × 10–6 C, ± 1 × 10–6 C (4) ± 3 × 10–6 C, ± 3 × 10–6 C
16. Ten positively charged particles are kept fixed on the x axis at points x = 10 cm, 20 cm, 30 cm, …, 100 cm. The first particle has a charge 1.0 ×1.0–8C, the second 8 × 10–8C, the third 27 × 10–8C and so on. Find the magnitude of the electric force acting on a 1 C charge placed at the origin.
(1) 2 × 105 N (approx) (2) 3 × 105 N (approx) (3) 4 × 105 N (approx) (4) 5 × 105 N (approx) 17. One brass plate is inserted between two charges.
The force between two charges will– (1) remain the same (2) increases (3) decrease (4) fluctuate 18. Electric lines of force can never start from –
(1) Proton (2) Conductor
(3) Insulator (4) Electron
19. For the following electric lines of forces, choose the correct option – A D B C (1) EB is maximum (2) EC is maximum (3) EA > ED (4) EB = EC
20. Two point charges exert on each other a force F when they are placed r distance apart in air. When they are placed R distance apart in a medium of dielectric constant K, they exert the same force. The distance R equals-(1) r K (2)
r
K
(3) rK (4)r K
21. Two charges q1 and q2 are placed in vacuum at a certain distance apart and the force acting between them is F. If a medium of dielectric constant 3 is introduced between them, then the force acting on
q1 becomes / remains– (1) 3 F (2) 3 2F (3) F (4) 3 F
22. What is dimensional formula of dielectric constant – (1) M
–1 L3 T–2A–2 (2) M–1 L–3 T4A2 (3) M0 L0 T0 (4) M L T–2
23. Dielectric constant of mica is –
(1) One (2) Less than one
(3) More than one (4) Infinite
24. Two point charges placed at a distance r in air exert a force F on each other. The value of distance R at which they experience force 4F when placed in a medium of dielectric constant K = 16 is–
(1) r (2) r/4
(3) r/8 (4) 2r
25. A proton and an electron are placed in a uniform electric
field-(1) the electric forces acting on them will be equal (2) the magnitudes of the forces will be equal (3) their accelerations will be equal
(4) the magnitudes of their accelerations will be equal 26. On putting salt [NaCl] in air a force F acts between
sodium and chlorine ions at a distance of 1 cm from each other. The permittivity of air and the dielectric constant of water are 0 and K respectively, when a piece of salt is placed in water then the force between Na+ and Cl–1 ions separated by a distance of 1 cm will
be-(1) F (2) 0 FK (3) 0 K F (4) K F
27. Two charges of opposite nature having magnitude 10 µC are 20 cm apart. The electric field at the centre of line joining these charges will
be-(1) 9 x 106 N/C in the direction of positive charge (2) 18 x 106 N/C in the direction of negative charge (3) 18 x 106 N/C in the direction of positive charge (4) 9 x 106 N/C in the direction of negative charge
28. Charge 2Q and –Q are placed as shown in figure. The point at which electric field intensity is zero will be–
–Q
A
+2Q B
(1) somewhere between –Q and 2Q (2) somewhere on the left of –Q (3) somewhere on the right of 2Q
(4) somewhere on the right bisector of line joining –Q and 2Q
29. Two point charges +1C and – 4C are placed at points having x co-ordinates x = a and x = 3a respectively on x-axis. Neutral point is
at-(1) x = 0 (2) x = –a
(3) x = 2a (4) x =
5
3
a
30. The electric field midway between two charges 0.1C and 0.4C separated by a distance of 60 cm is-(1) 5 103 N/C (2) 9 104 N/C
(3) 5 104 N/C (4) 3 104 N/C
31. Two point charges Q and –3Q are placed certain distance apart. If the electric field at the location of Q be E, then that at the location of –3Q will
be-(1) 3E (2) 3E
(3) E / 3 (4) E / 3
32. In the situation shown the string is ideal and insulating. Mass of the particle is 1mg, It’s charge will be (g = 10ms–2)
45o
E = 10 N/C2
(1) 1C (2) 10nC
33. 1C charge when placed at a point in electric field experiences 0.01N force. At the same point it 1mC charge is placed it will experience
force-(1) equal to 0.01N (2) equal to 10N (3) slightly less than 1N (4) slightly less than 10N
34. An isolated sphere of radius 1cm is placed in air maximum charge can be given to it without dielectric breakdown of the surrounding air
is-(1)
10
3
C
(2)100
3
nC
(3)100
3
C
(4)10
3
nC
35. How many electrons must be added to a spherical conductor of radius 10cm to produce a field of 2 10–3 N/C just above surface?
(1) 1.39 104 (2) 1.6 105
(3) 3 104 (4) 9.1 104
36. Which of the following graph best represents the variation of electric field intensity due to a charged solid sphere of copper?
(1) r O E (2) r O E (3) r O E (4) r O E
37. Along x axis at position x = a, x = 2a, x = 4a, x = 8a and so on, charges of strength q are placed. The electric field at origin will be – (1) 3 2 4 a Kq (2) 5 2 4 a Kq (3) 4 2 3 a Kq (4) 2 2 a Kq
38. A simple pendulum has a length l, mass of bob m. The bob is given a charge q coulomb. The pendulum is suspended in a uniform horizontal electric field of strength E as shown in figure, then calculate the time period of oscillation when the bob is slightly displace from its mean position is–
E l q m1 (1) g l 2 (2) m qE g l 2 (3) m qE g l 2 (4) 2 2 m qE g l 2
39. Three point charges q0 are placed at three corners of square of side a. Find out electric field intensity at the fourth corner.
(1) 2 0 a kq 2 1 2 (2) 20 a kq 2 1 2 (3) 2 0 a kq 2 1 2 (4) 2 0 a kq 2 1 1
40. Two insulating spheres of radii 2 cm and 4 cm have equal volume charge density. The ratio of electric field on the surfaces of the spheres will be –
(1) 1 : 2 (2) 4 : 1
41. The maximum electric field intensity on the axis of a uniformly charged ring of charge q and radius R will be– (1) 2 0 3 3R q 4 1 (2) 2 0 3R q 2 4 1 (3) 2 0 3 3R q 2 4 1 (4) 2 0 2 3R q 3 4 1
42 A circular ring of radius a carries a total charge Q distributed uniformly over its length. A small length dL of the wire is cut off. Find the electric field at the centre due to the remaining wire.
(1) 3 0 2 a 2 Qdl (2) 3 0 2 a 4 Qdl (3) 3 0 2 a 6 Qdl (4) 3 0 2 a 8 Qdl
43. A point charge q and a charge –q are placed at x = –a and x = +a respectively. Which of the following represents a part of E-x graph?
(1) E –a O +a x (2) E –a O x (3) E +a O x (4) all of these
44. A particle of mass m and charge q starts moving from rest along a straight line in an electric field E = E0 – ax where a is a positive constant and x is the distance from starting point. Find the distance travelled by the particle till the moment it came to instantaneous
rest-(1)
2
E
0a
(2)E
a
0 (3)E q
m
0 (4) E q045. A very long charged rod is placed along y = –x straight line as shown in figure, it carries charge per unit length. Electric field at point P is–
++ ++ ++ ++ ++ ++ ++ 45º y (m) x (m) (0, 6) P (1) 0 12 (2) 2 20 (3) 0 12 2 (4) 0 6
1. Dimensions of pE where p is electric dipole moment and E is electric field –
(1) MLT–2 (2) ML2T–2 (3) MLT–1 (4) ML–1T–2 2. If 1 E and 2
E are electric field at two points on equatorial line at distance r and 2r from short dipole, then E1/E2 – (1) 4 1 (2) 1 8 (3) 8 1 (4) 1 1
3. Due to an electric dipole shown in figure, the electric field intensity is parallel to dipole axis :
Y Q
–q +q P x
equitarial
(1) at P only (2) at Q only
(3) both at P and at Q (4) neither at P nor at Q 4. Three charges are arranged on the vertices of an equilateral triangle as shown in figure. Find the dipole moment of the combination.
–q 2q d –q (1) qd (2) qd 3 (3) 2qd (4) 2 2qd
5. An electric dipole is placed (not at infinity) in an electric field generated by a point charge–
(1) the net electric force on the dipole must be zero (2) the net electric forceon the dipole may be zero (3) the torque on the dipole due to the field must be zero
(4) the torque on the dipole due to the field may be zero
6. An electric dipole consists of two charges + q at a separation 2a. It is placed in such a way that it’s centre coincides with origin and dipole moment vector is directed towards + x axis. Electric field intensity magnitude at x = 2a is equal
to-(1)
kq
a
2
3 (2)2
3kq
a
(3)8
9
2kq
a
(4)kq
a
2
27. If an electric dipole is kept in a uniform electric field, then it will experience –
(1) Force only (2) Torque only (3) Force and torque (4) No force and no torque
8. An electric dipole is placed along the X-axis at the origin O. A point P is placed at a distance of 20 cm from this origin such that OP makes an angle
3
with the X-axis. If electric field at P makes an angle with X-axis, the value of
is-(1) 3 (2) tan 1 3 3 2 (3) 2 3 (4) 2 3 tan1
9. Find the magnitude of the electric field at the point P in the configuration shown in figure for d >> a. Take 2qa = p. P –q +q a +q a d (1) 3 2 2 0 p q d 4 1 (2) 2 2 2 3 0 p d q d 4 1 (3) 3 2 2 2 0 p d q d 4 1 (4) 2 2 2 0 p q d 4 1
EXERCISE – 1(B)
10. An electric dipole of moment p is placed at the origin along the x-axis. The angle made by electric field with x-axis at a point P, whose position vector makes an angle with x-axis, is (where tan = tan
2 1
)–
(1) (2)
(3) + (4) + 2
11. A point charge q and a dipole of dipole moment p are placed at some distance as shown in figure. The force on dipole is – q p x y r (1) 2 3 r kpq (2) 3 r kpq (3) 4 3 r kpq (4) 2 3 r kpq
12. An electric dipole having dipole moment 2 ´ 10–6 Cm is
placed in uniform electric field having magnitude 103
N/C. The dipole can experience maximum torque -(1) 2 × 10–3 Nm (2) 1 × 10–3Nm
(3) 4 × 10–3 Nm (4) zero
13. A small electric dipole of dipole moment p is perpendicular to electric field. Work done to rotate it through angle 180º in uniform electric field E is –
(1) pE (2) 2pE
(3) –2pE (4) Zero
14. An electric dipole consists of two opposite charges each of magnitude 1.0 C separated by a distance of 2.0 cm. The dipole is placed in an external field of 1.0 × 105 N/C. The maximum torque on the dipole is–
(1) 0.2 × 10–3 N-m (2) 1.0 × 10–3 N-m (3) 2.0 × 10–3 N-m (4) 4.0 × 10–3 N-m 15. The force between two short electric dipoles
separated by a distance r is directly proportional to–
(1) r2 (2) r4
(3) r–2 (4) r–4
16. A short electric dipole of dipole moment p is placed at origin with its dipole moment directedalong x-axis. The value of force experienced by a particle having charge –q0, placed at (0, a) is
(1) 3 0 0 2 4 1 a pq
along positive x-axis
(2) 3 0 0 2 4 1 a pq
along negative x-axis
(3) 3 0 0 4 1 a pq
along negative x-axis
(4) 3 0 4 1 a pq
along positive x-axis
1. Find out the electric flux through an area 10 m2 lying in XY plane due to an electric field
kˆ 5 jˆ 10 iˆ 2 E . (1) 25 Nm2/C (2) 50 Nm2/C (3) 75 Nm2/C (4) 100 Nm2/C
2. In a uniform electric field E if we consider an imagi-nary cubical close gaussian surface of side a, then find the net flux through the cube?
(1) 0 (2) Ea2
(3) 2 Ea2 (4) 6 Ea2
3. If electric field is uniform, then the electric lines of forces are–
(1) divergent (2) covergent (3) circular (4) paraller
4. A surface S = 10 is kept in an electric field E =jˆ kˆ
7 jˆ 4 iˆ
2 . How much electric flux will come out through the surface ?
(1) 40 unit (2) 50 unit
(3) 30 unit (4) 20 unit
5. In a region of uniform electric field E, a hemispheri-cal body is placed in such a way that field is parallel to its base (as shown in figure). The flux linked with the curved surface is–
–q O C E –q (1) zero (2) –R2E (3) R2E (4) E 2 R2
6. A square of side 'a' is lying in xy plane such that two of its sides are lying on the axis. If an electric field EE0xkˆ is applied on the square. The flux passing through the square is–
(1) 3 0a E (2) 2 a E0 3 (3) 3 a E0 3 (4) 2 a E0 2
7. Figure (a) shows an imaginary cube of edge L/2. A uniformly charged rod of length L moves towards left at a small but constant speed v. At t = 0, the left end just touches the centre of the face of the cube opposite it. Which of the graphs shown in figure (b) represents the flux of the electric filed throught the cube as the rod goes through it ?
v L L/2 (a) (b) flux a b c d time (1) a (2) b (3) c (4) d
8. The electric field in a region is given by E = ,
iˆ
ax where = constant of proper dimensions. What should be the charge contained inside a cube bounded by the surface, x = l, x = 2l, y = 0, y = l, z = 0, z = l ? (1) 3 0l (2) 3 0l (3) 30 l (4) 3 0l 2
9. If the electric flux entering and leaving a closed surface are respectively of magnitude 1 and 2, then the electric charge inside the surface will be– (1) 0 1 2 (2) (12)0 (3) 0(21) (4) 0(21)
10. Consider the charge configuration and a spherical Gaussian surface as shown in the figure. When calculating the flux of the electric field over the spherical surface, the electric field will be due to–
q2 + q1
– q1
(1) q2
(2) only the positive charges (3) all the charges
(4) + q1 and – q1
11. A point charge Q is placed at the centre of a hemishphere. The electric flux passing through flat surface of hemisphere is–
(1) 0 Q (2) zero (3) 0 2 Q (4) none of these
12. A point charge Q is placed at the cnetre of a hemisphere. the ratio of electric flux passing through curved surface and plane surface of the hemisphere is–
(1) 1 : 1 (2) 1 : 2 (3) 2 : 1 (4) 4 : 1 13. Select the wrong statement–
(1) the electric field calculated by Gauss's law is the field due to the charges inside the Gaussian surface
(2) the electric field calculated by Gauss's law is the resultant field due to all the charges inside and outside the closed surface (3) the Gauss's law is equivalent to Coulomb's
law
(4) the Gauss's law can also be applied to calculate gravitational field but with some modifications 14. A charge q is placed outside a hemisphere. Flux
through curved surface as shown in the figure–
q (1) 0 2 q (2) 0 2 q (3) Zero (4) 0 2 q
15. A point charge q is held, just below the centre of curvature of a hemispherical surface as shown in figure. The value of electric flux passing through the surface is (1) 0 q Centre of Curvature q (2) 0 2 q (3) More than 0 2 q
but less than 0 q (4) Less than 0 2 q
16. A charge Q is placed at the centre of an imaginary hemispherical surface. Using symmetry arguments and the Gauss's law, Find the flux of the electric field due to this charge through the surfaces of the hemisphere– Q (1) 0 Q (2) 2 0 Q (3) 0 Q 2 (4) 4 0 Q
17. A point charge Q is placed on the axis of a cone as shown in figure. If flux lined to curved surface is, then what is flux links to base of cone –
Q (1) 0 q (2) 0 q (3) Zero (4) 0 q
18. The electric field intensity at a distance r from an infinite sheet of charge with surface charge density is-(1) 0 (2) 0 2 (3) 0 2 2 (4) 2 r 0 2
19. If the electric field due to an infinite long line charge at distance 1m from it is 1 N/C, the charge per unit length of line charge is –
(1) 0 (2) 0
(3) 20 (4) 40
20. A point charge Q is placed at the centre of a cir-cular wire of radius R having charge q. The force of electrostatic interaction betwene poin charge and the wire is – q Q O (1) 2 0R 4 qQ (2) zero (3) R 4 q 0 2 (4) none of these
21. The electric flux passing through the sphere, if an electric dipole is placed at the centre of a sphere, is– (1) 0 1 (2) 0 2
(3) zero (4) none of these
22. Two parallel charged plates have a charge density of + and –. Net force on proton located outside the plates at some distance will be–
(1) e 0 (2) e 0 2 (3) e 0 2 (4) Zero
23. Electric charges are distributed in a small volume. The flux of the electric field through a spherical surface of radius 10 cm surrounding the total charge is 25 V-m. The flux over a concentric sphere of radius 20 cm will be–
(1) 25 V-m (2) 50 V-m (3) 100 V-m (4) 200 V-m
24. Eight point charges (can be assumed as small spheres uniformly charged and their centres at the corner of the cube) having values q each are fixed at vertices of a cube. The electric flux through square surface ABCD of the cube is–
A B C D q q q q q q q q (1) 0 24 q (2) 12 0 q (3) 0 6 q (4) 8 0 q
25. The charge density of a spherical charge distribution is given by:
r
n
r r n
r n
b g
F
HG
I
KJ
R
S|
T|
00
what is the total charge on the distribution? (1) 2 n3 0 (2) n 3 0 (3) 4/3 n3 0 (4) 1/2 n 3 0 26. T w o su rfa c e s S
1 and S2 enclose some charges as show in figure. What is ratio of flux linked to surfaces S1 and S2 – Q 2Q –2Q S1 S2 (1) 2 3 (2) 1 3 (3) 3 1 (4) 3 1
27. Two identical metal plates are having charges Q1 and Q2 (Q1 > Q2). They are placed at distance d, the electric field in between the plates is–
(1) Proportional to Q1 (2) Proportional to Q2 (3) Proportional to Q1 – Q2 (4) Proportional to Q1 + Q2
28. It a negatively charged particle is placed near an infinitely extended conducting plane, it will-(1) remain at rest there
(2) start moving parallel to plane (3) get repelled by the plane (4) get attracted by the plane
29. Charge Q is uniformly distributed over a triangle (equilateral) made from thin conducting wire each of length L. A gaussian sphere is taken with its centre at one vertex and passing through centroid of the triangle. Net electric flux linked with the sphere is-(1) Q 30 (2) 2 3 0 Q (3) 2 3 3 0 Q (4) zero
30. An electron moves along a metal tube with variable cross-section. The velocity of the electron when it approaches the neck of tube, is–
v0
(1) greater than v0 (2) equal to v0 (3) less than v0 (4) not defined
31. A square of side ‘a’ parallel to xy-plane is shown in figure. Co-ordinate of vertex P of square is (0, 0, a). A point charge qis placed at origin O then flux crossing the square is –
O x y R Q P S z a (1) 0 6 q (2) 0 24 q (3) 0 4 q (4) 0 2 q
32. Electric field at point P is given by E = rE0. The total flux through the given cylinder of radius R and height h is– P O r (1) E0R2h (2) 2E0R2h (3) 3E0R2h (4) 4E0R2h
33. A charge Q is placed at a distance of 4R above the centre of a disc of radius R. The magnitude of flux through the disc is . Now a hemispherical shell of radius R is placed over the disc such that it forms a closed surface. The flux through the curved surface taking direction of area vector along outward normal as positive, is–
4R Q R (1) zero (2) (3) – (4) 2
One or more correct choice type–
1. A point charge q is placed at origin. Let EA,EB and EC be the electric field at three points AA (1,2,3), B(1, 1, –1) and C(2,2,2) due to charge q. Then–
(1) EA EB (2) EA EC (3) EB4EC (4) EB8EC 2. A particle of charge q and mass m moves rectilinearly
under the action of an electric field E = – x. Here, and are positive constants and x is the distance from the point where the particle was initially at rest. Then–
(1) the motion of the particle is oscillatory
(2) the amplitude of the particle is
(3) the mean position of the particle is at x= (4) the maximum acceleration of the particle is
m q
3. A block of mass m is attached to a spring of force constant k. Charge on the block is q. A horizontal electric field E is acting in the direction as shown. Block is released with the spring in unstretched position–
E k
q, m
Smooth (1) block will execute SHM
(2) time period of oscillation is
k m 2 (3) amplitude of oscillation is k qE
(4) block will oscillate but not simple harmonically
4. A positively charged thin metal ring of radius R is fixed in the x-y plane with its centre at the origin O. A negatively charged particle P is released from rest at the point (0, 0, z0) where z0 > 0. Then the motion of P is–
(1) periodic for all values of z0 satisfying 0 < z0< .
(2) simple harmonic for all values of z0 satisfying 0 < z0 R
(3) approximately simple harmonic provided z0<<R (4) such that P crosses O and continues to move
along the negative z-axis towards z = – 5. A non- conducting solid sphere of radius R is
uni-formly charged. The magnitude of the electric field due to the sphere at a distance r from its centre– (1) increases as r increases for r < R
(2) decreases as r increases for 0 < r < (3) decreases as r increases for R < r < (4) is discontinuous at r = R
6. The electric field intensity at a point in space is equal in magnitude to–
(1) magnitude of the potential gradient there (2) the electric charge there
(3) the magnitude of the electric force, a unit charge would experience there
(4) the force, an electron would experience there 7. Figure shows a charge q placed at the centre of a
hemisphere. A second charge Q is placed at one of the positions A, B, C and D. In which position(s) of this second charge, the flux of the electric field through the hemisphere remains unchanged?
B C A D q (1) A (2) B (3) C (4) D
EXERCISE – 2
8. An electric dipole is placed at the centre of a sphere, mark the correct options.
(1) the flux of the electric field through the sphere is zero
(2) the electric field is zero at every point of the sphere
(3) the electric field is not zero anywhere on the sphere
(4) the electric field is zero on a circle on the sphere 9. An oil drop has a charge –9.6 × 10–19 C and has a mass 1.6 × 10–15 gm. When allowed to fall, due to air resistance force it attains a constant velocity. Then if a uniform electric field is to be applied vertically to make the oil drop ascend up with the same constant speed, which of the followign are correct (g = 10 ms-2) (Assume that the magnitude of resistance force is same in both the cases) (1) the electric field is directed upward (2) the electric field is directed downward
(3) the intensity of electric field is 102N.C 1 3
1
(4) the intensity of electric field is 105NC 1 6
1
10. An electric dipole is kept in the electric field pro-duced by a point charge.
(1) dipole will experience a force (2) dipole can experience a torque (3) dipole can be in stable equilibrium
(4) it is possible to find a path (not closed) in the field on which work required to move the dipole is zero.
COMPREHENSION BASED QUESTIONS I. Find out electric field intensity due to unfiormly
charged solid nonconducting sphere of volume
c h a rg e d e n sity and radius R at following points :
11. At a distance r from the surface of sphere
(inside)-(1) 3 r PR 0 (2) 3 rˆ ) r R ( P 0 (3) 0 R P (4) 3 0 R P
12. At a distance r from the surface of sphere (out- side)-(1) rˆ ) R r ( R P 2 0 3 (2) 2 (r R) rˆ R P 2 0 3 (3) 3 (r R) rˆ R P 2 0 3 (4) 4 (r R) rˆ R P 2 0 3
II. Two identical small spheres are suspended from the same point by threads 1m long. Each sphere is given a charge 120 nC. Consequently, they repel each other to a distance 40 cm.
13. Find the mass of each sphere–
(1) 400 mg (approx) (2) 500 mg (approx) (3) 600 mg (approx) (4) 700 mg (approx) 14. Find the tension in each thread–
(1) 2.1 × 10–3 N (2) 3.1 × 10–3 N (3) 4.1 × 10–3 N (4) 5.1 × 10–3 N
15. Find the angle between the two threads in the con-dition of equilibrium–
(1) sin–1(1) (2) sin–1(2) (3) sin–1(3) (4) sin–1(4)
III. A thin rod of length L carries a positive charge that is uniformly distributed over tis length. Linear charge density of the distribution, i.e. charge per unit length is . P is a point at a P 45° r Q A B O L d
distance ‘r’ from the midpoint of rod and on its perpendicular bisector. Q is a point along the axis of the rod and located at distance ‘d’ from one end. As shown, the line that joins P with any end of the rod makes an angle 45° with the perpendicular bi-sector. It appears that we could obtain electric field strength at various points with the help of Gauss’s
law. It is possible to consider a cylinder of radius r as the Gaussian surface with AB along its axis. Obviously, P will lie on the curved surface of this cylinder. But field strength at different points on the curved surface will be different owing to lack of symmetry of the distribution for various points. Thus Gauss’s law will not be practically useful. Similarly, we may imagine a cylinder with AB along its axis such that Q lies on one of its circular ends but Gauss’s law is again of little practical use. In such a situation we can obtain field strength by integrat-ing the field strength due to a small element usintegrat-ing suitable limits. (Take electric potential = 0, at ) 16. Let in figure point O be treated as origin and length
of rod along x axis so that P lies on y axis. x component of electric field strength at P will be–
(1) 2 r 0 (2) 2.83 r 0 (3) zero (4) r L 4 1 0
17. y-component of electric field strength at P is –
(1) 2 r 0 (2) 2.83 r 0 (3) zero (4) r L 4 1 0
18. Electric field strength at Q is (perhaps it could be more conveninent to shift the origin to Q for the purpose of field and potential calculation at Q)–
(1) 41 d(LLd) 0 (2) (L d) L 4 1 0 (3) (d L/2) L 4 1 0 (4) zero
Assertion and Reasons
(1) 1 is true, 2 is true; Statement-2 is a correct explanation for Statement-1.
(2) Statement-1 is true, Statement-2 is true; Statement-2 is not a correct explanation for Statement-1. (3) Statement-1 is true, Statement-2 is false. (4) Statement-1 is false, Statement-2 is true.
19. A : A point charge q is placed at centre of spherical cavity inside a spherical conductor as shown. An-other point charge Q is placed outside the conduc-tor as shown. Now as the point charge Q is pushed away from conductor, the potential difference (VA – VB) between two points A and B within the cavity of sphere remains constant.
q B
A Q
R : The electric field due to charges on outer sur-face of conductor and outside the conductor is zero at all points inside the conductor.
20. A : A solid uncharged conducting cylinder moves with acceleration a (w.r.t. ground). As a result of acceleration of cylinder, an electric field is produced within cylinder.
a solid conducting cylinder
R : When a solid conductor moves with accelera-tion a, then from frame of conductor a pseudoforce (of magnitude ma; where m is mass of electron) will act on free electrons in the conductor. As a result some portion of the surface of conductor acquires negative charge and remaining portion of surface of conductor acquires positive charge. 21. A : A charge q is placed at the center of a metallic
sheel as shown in figure. Electric field at point P on the shell due to charge q is zero.
P q
R : Net electric field in a conductor under electrostatic conditions is zero.
22. A : If there exists attraction between two bodies,
both of them may not be charged.
R : Due to induction effects a charged body can attract a neutral body.
23. A : When charges are shared between two bodies, there occurs no loss of charge, but there does occur a loss of energy.
R : In case of sharing of charges energy of conser-vation fails.
24. A : On going away from a point charge or a small electric dipole, electric field decrease at the same rate in both the cases.
R : Electric field is inversely proportional to square of distance from a point charge.
