It is clear from figure that pointP lies on the axial line of dipole with moment p1. Hence magnitude of the electric field intensity E 1
atP due to p1 is
Similarly,P lies on the equatorial line of dipole with moment p2.
Hence, magnitude of electric field intensity E 2
If the resultant field intensity vector E makes an angle with the direction of E 1,
16. DIPOLE IN A UNIFORM ELECTRIC FIELD 16.1 Torque
When a dipole is placed in a uniform field as shown in figure the net force on it
Electrostatics: Part 1 35
( ) 0
F R = È Î qE + - q E ˘˚=
While the torque t = qE l ¥ 2 sin i.e., q t =pE sinq {as p = q2l}
or t = p E ¥
From the expression it is clear that couple acting on a dipole is maximum (= pE ) when dipole is perpendicular to the field and minimum (= 0) when dipole is parallel or antiparallel to the field.
Illustration 25 Find the force on a small electric dipole of dipole moment p due to a point chargeQ placed at a distancer . Solution
Electric field of a point charge is a non-uniform electric field. Electric field at a distance x from the point charge is
2 3
0 0
1 1 2
4 4
Q dE Q
E pe x dx pe x
= fi =
-Magnitude of force on the dipole
3 0
1 2 x r 4
dE pQ
F p
dx = pe r
= =
Same can be calculated as force on the point charge due to dipole which is same as the force on dipole due to point charge (Newton’s 3rd law). The electric field of small dipole at a distancer is 3
0
1 2 4
E p
r
pe
= .
Hence force on the point chargeQ is 3
0
1 2 4
F pQ
r
pe
= .
Illustration 26 An electric dipole consists of two charges of 0.1 m C separated by a distance of 2.0 cm. The dipole is placed in an external field of 105 N/C. What maximum torque does the field exert on the dipole?
Solution
sin 2 sin
pE q a E
t = q = ¥ ¥ q
Max. value oft will be when sinq = 1 i.e.
\ t max = 10 - 7¥ ¥ 2 10 - 2¥ 10 5¥ = ¥1 2 10 N-m-4
Illustration 27 Two tiny spheres, each of mass M , and charges +q and –q respectively, are connected by a massless rod of length, L. They are placed in a uniform electric field at an angleq with the E (q ≈ 0o). Calculate the minimum time in which the system aligns itself parallel to the E
. Solution
t = pE sinq (asq → 0, sinq →q )
fi t = –( pE )q (If we assume angular displacement to be anti-clockwise, torque is clockwise)
fi E 0 2
I
a= - Ê Á Ë ˆ ˜ ¯ q= -w q p
As torque is proportional to 'q ' and oppositely directed, there will be an SHM Here, p =q.L and moment of inertia, I = M ( L /2)2 + M ( L /2)2 = ML2 /2
36 Electrostatics: Part 1
\ Time period T 2 I
pE
= p
The minimum time required to align itself is
4 T
16.2 Net Force on a Dipole in a Non-Uniform Field Suppose an electric dipole with dipole moment P
Æ
is placed in a non-uniform electric field E Æ=Eiˆ that points along x -axis. Let
E depend only on x . The electric field at the position of negative charge is E and at the position of positive charge ( E + DE ). Net force acting on the dipole is then
( )
F q E = + D - = DE qE q E
2 2 as
E dE E dE dE
q a aq p
x dx x dx dx
D D
È ˘ È ˘
= ÍÎD ˙ ˚= ÍÎ D = ˙˚ =
| | dE
F p
dx
Æ Æ
=
where dE
dx is the gradient of the field in the x -direction.
Concept Application Exercise 3
1. A point charge 50 µC is located in the x-y plane at the position vector r 0= (2 i j mˆ +3 ) .ˆ The electric field at the point of position vector r = (8 i j mˆ -5 ) ,ˆ in vector from is equal to:
(a) 90( 3 - i ˆ +4 ) V/ mˆj (b) 90(3 i ˆ -4 ) V/ mˆj (c) 900 ( 3 - i ˆ +4 ) V/ mˆj (d) 900(3 i ˆ -4 ) V/ mˆj
2. Two concentric rings, one of radius R and total charge +Q and the second of radius 2 R and total charge - 8 ,Q lie in x-y plane (i.e., z = 0 plane). The common centre of rings lies at origin and the common axis coincides with z-axis. The charge is uniformly distributed on both rings. At what distance from origin is the net electric field on z-axis zero.
(a)
2
R (b)
2 R
(c)
2 2
R (d) 2 R
3. A particle of mass 2 kg and charge 1 mC is projected vertically with a velocity 10 ms–1. There is a uniform horizontal electric field of 104 N/C.
(a) the horizontal range of the particle is 10 m (b) the time of flight of the particle is 2 s
(c) the maximum height reached is 5 m (d) the horizontal range of the particle is 0
4. A light weight particle of chargeQ is fixed at one end of an electrically insulated uniform elastic rod of natural length L, cross-sectional area A and Young’s modulusY . The rod is placed in space having uniform electric field of magnitude E and directed parallel to length of the rod as shown. Neglecting gravity, the magnitude of extension in this rod is
Electrostatics: Part 1 37
(a) QEL
YA (b)
2
QEL YA
(c)
4
QEL YA
(d) None of these
5. Consider two thin uniformly charged concentric shells of radiir and 2r having chargesQ
and –Q respectively, as shown. Three points A, B andC are marked at distances , 3 and 5
2 2 2
r r
respectively from their common centre. If E A, E B and E C are magnitudes of the electric fields at points A, B andC respectively then
(a) E A > E B > E C (b) E C > E B > E A (c) E B > E A = E C (d) E B > E A > Ec
6. If the magnitude of intensity of electric field at a distance x on axial line and at a distance y on equatorial line on a given dipole are equal, then x : y is
(a) 1 : 1 (b) 1 : 2
(c) 1 : 2 (d) 32 : 1
7. Three charges of (+2q), (–q) and (–q) are placed at the corners A, B andC of an equilateral triangle of side a as shown in the adjoining figure. Then the dipole moment of this combination is
(a) qa (b) Zero
(c) q a 3 (d) 2
3 qa
8. An electric dipole is placed along the x -axis at the originO. A pointP is at a distance of 20 cm from this origin such that
OP makes an angle
3
p with the x -axis. If the electric field atP makes an angleq with x -axis, the value ofq would be
(a)
3
p (b) tan 1 3
3 2
p - Ê ˆ
+ Á Ë ˜ ¯
(c) 2
3
p (d) tan 1 3
2
- Ê ˆ Á ˜ Ë ¯
9. An electric dipole in a uniform electric field experiences
(a) Force and torque both (b) Force but no torque
(c) Torque but no force (d) No force and no torque
10. A point charge placed at any point on the axis of an electric dipole at some large distance experiences a force F . The force acting on the point charge when its distance from the dipole is doubled is
38 Electrostatics: Part 1
(a) F (b)
2 F
(c)
4
F (d)
8 F
11. A point particle of mass M is attached to one end of a massless rigid non-conducting rod of leng th L. Another point particle of the same mass is attached to other end of the rod. The two particles carry
charges +q and –q respectively. This arrangement is held in a region of a uniform electric field E such that the rod makes a small angleq (say of about 5 degrees) with the field direction (see figure). What will be minimum time, needed for the rod to become parallel to the field after it is set free?
(a) 2
2
t mL
pE
= p (b)
2 2
t mL
qE
p
=
(c) 3
2 2
t mL
pE
p
= (d) t 2mL
qE
=p
12. An electric dipole is placed at the originO and is directed along the x -axis. At a pointP, far away from the dipole, the electric field is parallel to y-axis.OP makes an angleq with the x -axis then
(a) tan q = 3 (b) tan q = 2
(c) q = 45o (d) tan 1
2 q =
13. Two non-conducting spherical shells are uniformly charged. One shellS 1 having radius 5 m and charge –2mC has centre at (–1, 2, –1) and the other shellS 2 has radius 6 m and charge 3mC and centre at (–2, 1, –1). Find the electric field (in vector form) at point (1, –1, 3).
14. Two point charges are placed at pointa andb. The field strength to the right of the chargeQb
on the line that passes through the two charges varies according to a law that is represented graphically in the figure. Find the signs of the charges and ratio of magnitudes of charges
a b
Q Q
and the distance X 2 of the point fromb where the field is maximum, in terms of l and X 1. The electric field is taken positive if its direction is towards right and negative if its direction is towards left.
15. A square loop of side ‘l ’ having uniform linear charge density ‘l ’ is placed in ‘ xy’ plane as shown in the figure. There is a non-uniform electric field E a( x l i)ˆ
= l +
wherea is a constant. Find the resultant electric force inm N on the loop ifl = 10 cm,
a = 2 N/C and charge densityl = 2mC/m.
Electrostatics: Part 1 39