Department of PhysicsIIT Kanpur, Semester II, 2015-16

PHY103A: Physics II Solution # 4 Instructors: AKJ & DC

Solution 4.1: Force with image charges (Griffiths 3rd ed. Prob 3.6)

As far as force is concerned, this problem is the same if we remove the grounded conducting plate and simply putan image charge +2q at z = −d and an image charge −q at z = −3d. Therefore, the force on the charge +q is givenby

F =1

4πε0

[ −2q

(2d)2+

2q

(4d)2+

−q

(6d)2

]z =

q2

4πε0d2

[−12

+18

+−136

]z = − q2

4πε0d2

2972

z

Solution 4.2: Infinite-line image charge (Griffiths 3rd ed. Prob 3.9)

This problem is an extension of the problem in which a point charge is placed above an infinite grounded conductingplane. So, in order to obtain the correct potential, we need to put an infinite line charge −λ running parallel to thex-axis at a distance −d from the x− y plane. (see Fig. 1).

FIG. 1:

(a) Let’s first find the potential at a point (y, z) on the y − z plane. Assume that the distance of the point (y, z)from the top infinite line charge is given by the magnitude of the vector s+; similarly the distance of the point(y, z) from the bottom infinite line charge is given by the magnitude of the vector s−. Taking the potentialV = 0 at x − y plane to be the reference, the potential due to top infinite line charge can be shown to be

V+ = − 2λ

4πε0ln

(s+

d

). Similarly, the potential due to the bottom infinite line charge can be shown to be

V− = +2λ

4πε0ln

(s−d

). Therefore the total potential at (y, z) is given by

V (y, z) = − 2λ

4πε0ln

(s+

d

)+

2λ

4πε0ln

(s−d

)

=2λ

4πε0ln

(s−s+

)

=λ

4πε0ln

(s2−

s2+

)

=λ

4πε0ln

[y2 + (z + d)2

y2 + (z − d)2

]

1

(b) The induced charge density is given by σ = −ε0∂V

∂n, where n is the vector in the direction perpendicular to

the surface. In the present problem, n = z. Therefore,

σ = −ε0∂V

∂n= −ε0

∂V

∂z

∣∣∣z=0

= −ε0∂

∂z

[λ

4πε0ln

[y2 + (z + d)2

y2 + (z − d)2

]] ∣∣∣z=0

= −ε0λ

4πε0

[2(z + d)

y2 + (z + d)2− 2(z − d)

y2 + (z − d)2

] ∣∣∣z=0

= −ε0λ

4πε0

[2d

y2 + z2+

2d

y2 + d2

]

= − λd

π(y2 + d2)2

Solution 4.3: Far-field potential (Griffiths 3rd ed. Prob 3.26)

In this problem, a point r on the sphere is represented in the (r, θ, φ) coordinates. An observation point r0 isrepresented in the (r0, θ0, φ0) coordinates. The angle between the vectors r and r0 is represented by α. So, theformula for multipole expansion of potential V can be written as

V (r0) =1

4πε0

∞∑n=0

1rn+10

∫rnPn(cos α)ρ(r)dτ

Now, since we are interested in observation points along z-axis only, the angle α between the two vectors in essentiallythe angle θ. Therefore we can write the above expression as

V (r0) =1

4πε0

∞∑n=0

1zn+1

∫rnPn(cos θ)ρ(r)dτ

Let’s calculate the above potential term by term in the limit z À R.

1. Monopole term: We have

Vmono(r0) =1

4πε0z

∫ρ(r)dτ

=1

4πε0z

∫ [k

R

r2(R− 2r) sin θ

]r2 sin θdrdθdφ

=1

4πε0zkR

∫(R− 2r) sin2 θdrdθdφ

The r-integral yields∫ R

0(R− 2r)dr = (Rr − r2)

∣∣∣R

0= R2 −R2 = 0. Therefore, the monopole potential is zero:

Vmono(r0) =1

4πε0zkR

∫(R− 2r) sin2 θdrdθdφ = 0

2

2. Dipole term: We have

Vdip(r0) =1

4πε0z2

∫rP1(cos θ)ρ(r)dτ

=1

4πε0z2

∫r cos θρ(r)dτ

=1

4πε0z2

∫r cos θ

[k

R

r2(R− 2r) sin θ

]r2 sin θdrdθdφ

The θ-integral yields∫ π

0sin2 θ cos θdθ =

sin3 θ

3

∣∣∣π

0=

13(0− 0) = 0. Therefore, the dipole potential is also zero:

Vdip(r0) =1

4πε0z2

∫r cos θ

[k

R

r2(R− 2r) sin θ

]r2 sin θdrdθdφ = 0

3. Quadrupole term: We have

Vquad(r0) =1

4πε0z3

∫r2P2(cos θ)ρ(r)dτ

=1

4πε0z3

∫r2(3 cos2 θ − 1)/2ρ(r)dτ

=1

8πε0z3

∫r2(3 cos2 θ − 1)

[k

R

r2(R− 2r) sin θ

]r2 sin θdrdθdφ

The three integrals can be calculated as follows:

r − integral :∫ R

0

r2(R− 2r)dr =(

r3

3R− r4

2

) ∣∣∣R

0= −R4

6

θ − integral :∫ π

0

(3 cos2 θ − 1) sin2 θdθ = 2∫ π

0

sin2 θdθ − 3∫ π

0

sin4 θdθ = 2(π

2

)− 3

(3π

8

)= π

(1− 9

8

)= −π

8

φ− integral :∫ π

0

dφ = 2π

Therefore,

Vquad(r0) =kR

8πε0z3

(−R4

6

)×

(−π

8

)× (2π) =

14πε0

kπ2R5

48z3

So, the approximate potential is the potential of the quadrupole:

V (r0) ≈ Vquad(r0) =1

4πε0

kπ2R5

48z3

Solution 4.4: Potential due to a four-charge system (Griffiths 3rd ed. Prob 3.27)

The total charge of the system is zero. So, the monopole term in the potential would be zero. Now, let’s calculatethe dipole term [see Fig. 2]. We have

p = (3qa− qa)z + (−2qa− 2q(−a))y = 2qaz

Therefore,

V ≈ Vdipole =1

4πε0

p · rr2

=1

4πε0

2qaz · rr2

=1

4πε0

2qa cos θ

r2

3

y

x

z

+3q

-2q -2q

+q

a

a

aa

r

µ

FIG. 2:

Solution 4.5: Far-field Potential due to a spherical-charge distribution (Griffiths 3rd ed. Prob 3.28)

(a) From the symmetry in the problem it is clear that the net dipole moment will be in the z direction. Therefore

p ≡∫

rρ(r)dτ =∫

zzρ(r)dτ =∫

zρ(r)dτ z =∫

zσdaz

=∫

(R cos θ)(k cos θ)R2 sin θdθdφz

= 2πR3k

∫ π

0

cos2 θ sin θdθz

= 2πR3k

(−cos3 θ

3

) ∣∣∣π

0z

=4πR3k

3z

(b) We find that the total charge of the system is zero. So, there will be no monopole contribution to the far-fieldpotential. However since the dipole moment is not zero, the first contribution will be from the dipole term andthe far-field potential can be written as

V ≈ Vdipole =1

4πε0

p · rr2

=1

4πε0

4πR3k

3z · rr2

=1

4πε0

4πR3k

3cos θ

r2=

kR3

3ε0

cos θ

r2

Solution 4.6: Electric field of a pure dipole (Griffiths 3rd ed. Prob 3.33)

The electric field of a pure dipole is given by

Edip(r) =p

4πε0r3(2 cos θr + sin θθ)

=1

4πε0r3(3p cos θr− p cos θr + p sin θθ)

=1

4πε0r3[3(p · r)r− (p · r)r− (p · θ)θ]

=1

4πε0

1r3

[3(p · r)r− p]

Solution 4.7: Field and potential due to a three-charge system (Griffiths 3rd ed. Prob 3.32)

4

The total charge of the three-charge system is −q. Therefore the monopole contribution is not zero and is given by

Vmono =1

4πε0

−q

r= − q

4πε0r

The dipole moment p of the charge system is p = qaz + [−qa − q(−a)]y = qaz. So, the dipole contribution to thepotential is

Vdip =1

4πε0

p · rr2

=1

4πε0

qa cos θ

r2

Therefore, the potential up to two terms in the multipole expansion is given by

V ≈ Vmono + Vdip =q

4πε0

(−1

r+

qa cos θ

r2

)

The electric field E is given by

E = −∇V ≈ q

4πε0

[− 1

r2r +

a

r3

(2 cos θr + sin θθ

)]

y

x

z

+q

-q -qa

a

a

r

FIG. 3:

5