20.1 Identification of the Relevant Concepts
Gauss’s law is most useful in situations where the charge distribution has spherical or cylindrical symmetry or is distributed uniformly over a plane. In these situations we determine the direction of E Æ from the symmetry of the charge distribution. If we are given the charge distribution, we can use Gauss’ law to find the magnitude of E Æ. Alternatively, if we are given the field, we can use Gauss’s law to determine the details of the charge distribution. In either case, begin your analysis by asking the question, “What is the symmetry?”
The problem is solved by using the following steps:
(a) Select the surface that you will use with Gauss’s law. We often call it a Gaussian surface. If you are trying to find the field at a particular point, then that point must lie on your Gaussian surface.
Charge distribution Gaussian surface Electric field
Point charge
Spherical charge distribution Line of charge
Planer charge
Spherical Spherical Cylindrical
Plane parallel to charge distribution
Radial Radial Radial
Normal to surface
48 Electrostatics: Part 1
(b) The Gaussian surface does not have to be a real physical surface, such as a surface of a solid body. Often the appropriate surface is an imaginary geometric surface; it may be in empty space, embedded in a solid body, or both.
(c) Usually you can evaluate the integral in Gauss’ law (without using a computer) only if the Gaussian surface and the charge distribution have some symmetry property. If the charge distribution has cylindrical or spherical symmetry, choose the Gaussian surface to be a coaxial cylinder or a concentric sphere, respectively.
Illustration 37 Careful measurement of the electric field at the surface of a black box indicates that the net outward flux through the surface of the box is 8.0 × 103 Nm2 /C.
(a) What is the net charge inside the box?
(b) If the net outward flux through the surface of the box were zero, could you conclude that there were no charges inside the
box? Why or why not? [NCERT]
Solution
Here: 8.0 10 Nm /c3 2
f c = ¥
(a) As
2 2
12 3 6
0 2
0
C Nm
, 8.85 10 8.0 10 0.07 10 0.07 C
Nm C
c c
q q C
f f Ê - ˆÊ ˆ
-= Œ =Œ =ÁË ¥ ˜¯ÁË ¥ ˜¯ = ¥ = m
(b) Whenf c = 0,q = 0, whereq is the net charge inside the box. Thus, there may be charges inside the box whose algebraic sum is zero. Hence, we can only conclude that the net charge inside the box is zero.
Illustration 38 A point charge causes an electric flux of –1.0 × 103 Nm2 /C to pass through a spherical Gaussian surface of 10.0 cm radius centred on the charge.
(a) If the radius of the Gaussian surface were doubled, how much flux would pass through the surface?
(b) What is the value of the point charge? [NCERT]
Solution
Here, 1.0 10 Nm /C3 2
f c = - ¥
r = 10.0 cm = 10 × 10–2 m = 10–1 m
(a) As
0 c , f = q
Œ
it is clear thatf c depends onq only. Thus, even when the radius of the Gaussian surface is doubled, q remains the same and so isf c, i.e.,f c = –1.0 × 103 Nm2 / C
(b) q=Œ0f c =(8.85 10 ¥ -12C /Nm ) ( 1.0 10 N m /C) 2 2 - ¥ 3 2
8.85 10 -9 C 8.85 nC
= - ¥ =
-20.2 Field of a Charged Conducting of Sphere
We place positive chargeq on a solid conducting sphere with radius R (as shown in figure). All the charges must be on the surface of the sphere.
20.3 Selection of Gaussian Surface
The system has spherical symmetry. To take advantage of the symmetry, we take as our Gaussian surface an imaginary sphere of radiusr centred on the conductor. To calculate the field outside the conductor, we taker to be greater than the conductor’s radius R; to calculate field inside, we taker to be less than R. In either case, the point where we want to calculate E lies on Gaussian surface.
Electrostatics: Part 1 49
20.4 Electric Field Outside the Sphere
We first consider the field outside the conductor, so we chooser > R. The entire conductor is within the Gaussian surface, so the enclosed charge isq. The area of the Gaussian surface is 4p r 2; E is uniform over the surface and perpendicular to it at each point.
The flux integral
Ú
E dA^ in Gauss’s law is therefore just E (4p r 2) gives2
2
0 0
(4 ) and 1 (outside a charged conducting sphere) 4
q q
E r E
r
p
e pe
= =
This expression for the field at any point outside the sphere (r > R) is the same as for a point charge; the field due to the charged sphere is the same as though the entire charge were concentrated at its centre. Just outside the surface of the sphere, where r = R,
2 0
1 (at the surface of a charged conducting sphere) 4
E q
R
pe
=
20.4.1 Field of a Line Charge
Electric charge is distributed uniformly along an infinitely long, thin wire. The charge per unit length isl (assumed positive).
The system has cylindrical symmetry. This property suggests that we use as a Gaussian surface a cylinder with arbitrary radiusr and arbitrary lengthl, with its ends perpendicular to the wire. We break the surface integral for the fluxF E into an integral over each flat end and one over the curved side walls.
There is no flux through the ends because E lies in the plane of the surface.
To find the flux through the side walls, note that E is perpendicular to the surface at each point; by symmetry, E has the same value everywhere on the walls. The area
of the side walls is 2p rl. (To make a paper cylinder with radiusr and heightl, you need a paper rectangle with width 2p r , heightl, and area 2p rl). Hence the total fluxF E through the entire cylinder is the sum of the flux through the side walls, which is ( E ) (2p rl) and the zero flux through the two ends.
Finally, we need the total enclosed charge, which is the charge per unit length multiplied by the length of wire inside the Gaussian surface, orQenel=l l. From Gauss’s law,
0 0
( )(2 ) and 1
E 2
E rl l E
r
l l
p e pe
F = = =
(field of an infinite line of charge).
50 Electrostatics: Part 1
We have assumed thatl is positive. If it is negative, E is directed radially inward toward the line of charge, and in the above expression for the field magnitude E we must interpretl as the magnitude (absolute value) of the charge per unit length.
20.4.2 Field of an Infinite Plane Sheet of Charge
Let us consider a thin, flat, infinite sheet on which there is a uniform positive charge per unit areas .
To take advantage of these symmetry properties, we use as our Gaussian surface a cylinder with its axis perpendicular to the sheet of charge, with ends of area A.
The charged sheet passes through the middle of the cylinder’s length, so the cylinder, is perpendicular to the surface; hence, the flux through each end. Because E is perpendicular to the charged sheet, it is parallel to the curved side walls of the cylinder, and there is no flux through these walls.
The total flux integral in Gauss’s law is then 2 EA ( EA from each end and zero from the side walls). The net charge within the Gaussian surface is the charge per unit area multiplied by the sheet area enclosed by the surface, or Qenel =s A. Hence, Gauss’s law gives
0 0
2 and (field of an infinite sheet of charge) 2
EA s A E s
e e
= =
If the charge density is negative, E is directed toward the sheet, the flux through the Gaussian surface in figure is negative, and in the expression
2 0
E s e
= denotes the magnitude (absolute value) of the charge density.
The assumption that the sheet is infinitely large is an idealization; nothing in nature is really infinitely large. But the result
2 0
E s e
= is a good approximation for points that are close to the sheet (compared to the sheet’s dimensions) and not too near its edges. At such points, field is very nearly uniform and perpendicular to plane.
20.5 Field at the Surface of a Conductor
To find a relation betweens at any point on the surface and the perpendicular component of the electric field at that point, we construct a Gaussian surface in the form of a small cylinder (as shown in figure).
One end face, with area A, lies within the conductor and the other lies just outside.
The electric field is zero at all points within the conductor. Outside the conductor the component of E perpendicular to the side walls of the cylinder is zero, and over the end face the perpendicular component is equal to E ^.
(Ifs is positive, the electric field points out of the conductor and E ^ is positive; if s is negative, the field points inward and E ^ is negative.) Hence the total flux through the surface is E ^ A. The charge enclosed within the Gaussian surface is s A, so from Gauss’s law,
0 0
and (field at the surface of a conductor)
E A s A E s
e e
^ = ^ =
We can check this with the results we have obtained for spherical, cylindrical and plane surfaces.
20.5.1 Field of a Uniformly Charged Sphere
Positive electric chargeQ is distributed uniformly throughout the volume of an insulating sphere with radius R. 20.6 Electric Field Inside the Sphere
From symmetry the magnitude Eof the electric field has the same value at every point on the Gaussian surface, and the direction of
E
is radial at every point on the surface. Hence, the total electric flux through the Gaussian surface is the product of E and the total area of the surface A = 4p r 2, that is,F E = 4p r 2 E .
Electrostatics: Part 1 51
The amount of charge enclosed within the Gaussian surface depends on the radius r . Let’s first find the field magnitude inside the charged sphere of radius R; the magnitude E
is evaluated at the radius of the Gaussian surface, so we chooser< R.
The volume charge densityr is the chargeQdivided by volume of the entire charged sphere of radius R:
3 3
encl encl
3 3 3
; 4
4 3 4 /3 3
Q Q r
Q V r Q
R R R
r r p
p p
Ê ˆ Ê ˆ
= = =Á Ë ˜ ¯ Á ˜ Ë ¯ =
Then using Gauss’s law, enc.
0
E ds Q f =
Ú
◊ = e2 3
3 3
0 0
4 or 1 (field inside a uniformly charged sphere) 4
Q r Qr
ds E r E
R R
E p
e pe
= ◊ = =
Ú
The field magnitude is proportional to the distancer of the field point from the centre of the sphere.
At the centre (r = 0), E = 0.
Electric field in terms of charge density
3
3
0 0 0
4
1 3
(field inside a uniformaly charged sphere)
4 3 3
R r
r r
E E
R r p
r r
pe e e
Ê ◊ ˆ
Á ˜
Ë ¯
= = fi =
To find the field magnitude outside the charged sphere, we use a spherical Gaussian surface of radiusr > R. This surface encloses the entire charged sphere, soQenel =Q and Gauss’s law gives
2
0 0 2
4 or 1 (field outside a uniformly charged sphere) 4
Q Q
E r E
r
p
e pe
◊ = =
For any spherically symmetric charged body the electric field outside the body is the same as though the entire charges were concentrated at the centre.
20.7 Electric Field Due to a Long Uniformally Charged Cylinder
Consider a long uniformally charge cylinder of volumetric charge densityr and radius R.
52 Electrostatics: Part 1
For any pointr < R orr > R the Gaussian surface will be cylindrical as shown in figure.
For any point inside the cylinder (r < R).
2 in
0 0
( )
(2 ) q r l
E p r l r p
e e
◊
◊ = =
2 0
E r r E r e
= fi µ
For any point outside the cylinder (r > R)
2 in
0 0
( )
(2 ) q R l
E p r l r p
e e
◊
◊ = =
2
0
1 2
E R E
r r
r e
= fi µ
Electric field inside the long uniformally charged cylinder varies linearly E µ r
and outside the cylinder the electric field varies inversely proportional to the distance from the axis Ê Á Ë E µ1r ˆ ˜ ¯ .
Illustration 39 Consider two concentric conducting spheres. The outer sphere is hollow and initially has a charge –7Q on it.
The inner sphere is solid and has a charge +2Q on it.
How much charge is on the outer surface and inner surface of the outer sphere.
If a wire is connected between the inner and outer spheres, after electrostatic equilibrium is established, how much total charge is on the outer sphere? How much charge is on the outer surface and inner surface of outer sphere? Does the electric field at the surface of the inside sphere change when the wire is connected?
We return to original condition in (a). We now connect the outer sphere to ground will be on the outer sphere? How much charge will be on the inner surface and outer surface of the outer sphere?
Solution
(a) The charge on the inner sphere induce equal magnitude of charge, but opposite in sign, on the inner surface of outer sphere. Sum of all the i nduced charged is always zero. Therefore, an equal amount of charge must come on the outer surface. Thus outer and inner surface of outer sphere have charges –5Q and –2Q respectively.
(b) When outer and inner spheres are connected by a wire, the entire charge is transferred to the outer sphere from inner sphere. In electrostatic equilibrium charge does not reside inside a conductor.
Total charge on outer surface of outer sphere is –5Q. Total charge on inner surface is 0.
The electric field at the surface of the inside sphere goes to zero after connection.
Consider a Gaussian surface just on the surface of inner sphere.
2 enclosed
enclosed 0
(4 ) 0, as 0
E
E r Q Q
f p
e
= = = =
Thus, we have E = 0.
(c) When the outer sphere is grounded the charge on this surface is transferred to ground, thus charge is reduced to zero. The final charge distribution is shown in figure.
Electrostatics: Part 1 53
Illustration 40 A cube of side l has one corner at the origin of coordinates and extends along the positive x , y and z-axes.
Suppose the electric field in this region is given by E = ( a by j+ ) .ˆ Determine the charge inside the cube.
Solution
The faces adfe, bcgf , cdgh, adhe will contribute zero flux because the area vector is normal to electric field.
Flux through faceefgh,
2 2
2 E dA a j l ( ) ( )ˆ ˆj al f
Æ Æ
= ◊ =
Ú
◊ =-As the field at the face efgh (that lies in the yz plane, y = 0) is E aj= ˆ and area vector is
2( ˆ)
l -j (direction outward normal)
Flux through faceabcd f =(a bl j l j + )ˆ ◊ 2 ˆ=(al 2 +bl3)
Net flux through the cube = f 1 +f 2 =bl3
From Gauss’s law enclosed enclosed 0 0 3
0
. Hence, E
Q Q bl
f e f e
e
= = =
Illustration 41 A point chargeq is placed at a distance
2
a from the centre of a square of side ‘a’ as shown in the diagram;
calculate the electric flux passing through the square.
Solution
• This problem can solved by symmetry consideration and Gauss law.
• We can enclose the charged particle by a cube of side ‘a’ and keeping the particle at the centre of the cube.
• The total flux passing through the close cube
0
q .
f =e
• All the six surfaces are symmetrical with respect to charge, hence they will have equal contribution of the flux
0
. 6 6
f q
f ¢= = e
Illustration 42 In figure shown a chargeq is placed at a distanced Æ 0 near one of the edges of a cube of edge l on a line of symmetry along diagonal. What is flux through each of the sides containing the point a. What is the flux through the other three faces?
Solution
• Use of symmetry consideration may be useful in problems of flux calculation.
• We can imagine a charged particle is placed at the centre of a cube of side 2l.
• The flux enclosed with the cube
0
q .
f = e
54 Electrostatics: Part 1
• The flux passing through one of the face of the cube;
0
. 6 6
f q
f ¢= = e
• Hence the flux passing through the face
0
. 4 24
bcfg f q e
= ¢ =
• Each of the face (efgh), (bcfg) and (dchg) are symmetrical with respect to change hence the flux passing through each of the face is
0
. 24
q e
• The electric field lines for the faces (efgh), (bcfg) and (dchg) are away from the faces hence the flux associated with each of the faces will be positive
0
i.e. .
24 q
e
+ ˆ ˜ ¯
• Hence total flux through these side
0 0
3 .
24 8
q q
e e
= ¥ +
• Asd Æ 0 we can say the faces (abcd ). (abef ) and (adeh) are also symmetrical about charge.
• The number of electric field lines which are passing through the sides which do not contain the poin t a are same as the number of the electric field lines passing through the sides containing the point a.
• Hence same amount of the flux will pass through the sides contain the pointa , i.e.
0
. 8
q e
• The electric field lines are towards the faces containing the point a hence the flux will be negative i.e.
0
. 8
f q
= - e
¢¢
• Hence the flux through each of the faces containing the point ‘a’ will be
0
.
3 24
f q
e
¢¢=
- Illustration 43 Two identical metal plates each having surface area ‘ A’, having charge ‘q1’ and ‘q2’ are placed facing each other at a separation ‘d ’. Find the charge appearing on surface (1), (2), (3) and (4). Assume the size of the plate is much longer than the separation between the plates.
Electrostatics: Part 1 55
Solution
Facing surfaces have equal and opposite charge (By Gauss theorem). Let the facing surfaces have the charge x and – x [surfaces (2) and (3) respectively]. Then the charge on the surfaces (1) and (2) should be (q1 – x ) and x respectively (by conservation of charge). Facing metallic surfaces always have equal and opposite charge hence charge appearing on surfaces (3) and (4) will be – x and (q2 + x ) respectively. Let us consider a point ‘P’ inside the left plate. Net electric field at ‘P’ should be zero.
Net electric field at ‘P’ will be due to the resultant of electric field due to charge appearing on all four surfaces.
1 2 3 4
Hence charge appearing on different surfaces are as follows:
3 4
56 Electrostatics: Part 1
Note:
• Facing surfaces have the equal and opposite nature charge with magnitude ‘half the difference of the charge on different plates’, i.e., 1 2
2 q -q
Ê ˆ
Á ˜
Ë ¯ in surface (2) and ( 2 1) or 1 2
2 2
q - q Ê q - q ˆ
- Á Ë ˜ ¯ in surface (3).
• Outer surfaces always have equal charges of magnitude ‘half the summation of charges, i.e., ( 1 2)
2
q +q
in each surfaces (1) and (4).
• If we have this type of charge distribution then the electric field inside any metal plate will be zero.
• The charge appearing on the surfaces (2) and (3) is called bounded charge and the charge appearing on the surfaces (1) and (4) is called free charge.
• If we join second plate (right plate) with ground the charge appearing on the surface (4) will go to the earth and charge distribution will be
• Any metal plate or object connected to the earth need not to have zero charge. If conducting body is isolated and connected to earth it will have no charge. If the conducting body is connected to earth have any charged object near to it the body will not have zero charge.
21. APPENDIX
Solid angle: It is the cone subtended by an area at the point of interest. The magnitude of solid angle subtended by an area S at a point is define as
2
cos
S
ds r
W =
Ú
qat its centre. Therefore, total solid angle around a point in space is the solid angle subtended by entire spherical surface on its centre.
2
0 2
4 R 4 steradian R
p
W = = p
Solid angle subtended by a disk at a point on its axis: Consider a coaxial area element of radius x and thicknessdx .
dS = 2p xdx
Solid angle subtended by this element at pointP is
2 2 2 2 2 2
cos 2
( ) ( )
dS xdxa
d d
x a x a a x
q p
W = fi W =
+ + +
Hence total solid angle subtended by the disk is
2 2 3/2 2 2 2 2
0 0
2 1
2 2 1
( )
R R
xdx a
a a
x a a x a R
p p p
Ê ˆ Ê ˆ
W =
Ú
+ fi W = ÁË- + ˜¯ fi W = ËÁ - + ˜¯W = 2p (1– cosa ); wherea is the semi vertical angle of the cone subtended by the disk atP.
Electrostatics: Part 1 57
Illustration 44 A point chargeq is placed on the apex of a cone of semi-vertex angle q . Show that the electric flux through the base of the cone is
0
Method 1: For point charge Gaussian surface should be spherical. Consider a Gaussian sphere with its centre at the apex and radius the slant length of the cone. The flux through the whole sphere isq / e 0. Therefore, the flux through the base of the
cone, E 0 0
and A = area of sphere below the base of the cone.
Consider a differential ring of radiusr and thicknessdr .
dA = (2p r ) Rd a = (2p Rsina ) R sina ) Rd a asr = Rsin a =(2p R2) sina d a
Method 2: Using the concept of solid angle.
Solid angle is the cone subtended by an arc at the point of intersect. Total slid angle around a point in space is 4l steradian. Solid angle subtended by the base of the cone at the apex of cone isp = 2p (1 – cosq ). As the flux associated with solid angle 4p is
0
q . e
Hence the flux associated with solid angle
0
1. The inward and outward electric flux for a closed surface in units of N-m2 /C are respectively 8 × 103 and 4 × 103. Then the total charge inside the surface is [wheree 0= permittivity constant]
(a) 4 × 103 C (b) –4 × 103 C
(c) ( 4 10 )3
C e
- ¥ (d) –4 × 103e 0 C
2. A chargeq is placed at the centre of a cube. Then the flux passing through one face of cube will be
(a)
3. If a spherical conductor comes out from the closed surface of the sphere then total flux emitted from the surface will be
(a)
0
1 e
¥ (the charge enclosed by surface) (b) e 0 × (charge enclosed by surface)
(c)
0
1 4pe
¥ (charge enclosed by surface) (d) 0
58 Electrostatics: Part 1
4. A chargeq is located at the centre of a cube. The electric flux through any face is
(a)
0
4 6(4 )
p q
pe (b)
6(4 0)
p q pe
(c)
6(4 0)
q pe
(d)
0
2 6(4 )
p q pe
5. Shown below is a distribution of charges. The flux of electric field due to these charges through the surfaceS is
(a) 3q / e 0 (b) 2q / e 0
(c) q / e 0 (d) Zero
6. Consider the charge configuration and 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
(a) q2 (b) Only the positive charges
(c) All the charges (d) +q1 and –q2
7. An electric dipole is put in north-south direction in a sphere filled with water. Which statement is correct?
(a) Electric flux is coming towards sphere
(b) Electric flux is coming out of sphere
(c) Electric flux entering into sphere and leaving the sphere are same
(d) Water does not permit electric flux to enter into sphere
8. The electric flux for Gaussian surface A that enclose the charged particles in free space is (givenq1 = –14 nC,q2 = 78.85 nC,
8. The electric flux for Gaussian surface A that enclose the charged particles in free space is (givenq1 = –14 nC,q2 = 78.85 nC,