Cornell University, Physics Department Fall 2004PHYS-217 Electricity & Magnetism Section 01

Solutions to Problem Set 2

David C. Tsang

Purcell §1.10

At the beginning of the century the idea that the rest mass of the electron mighthave a purely electrical origin was very attractive, especially when the equivalenceof energy and mass was revealed by special relativity. Imagine the electron as aball of charge, of constant volume density out to some maximum radiusro. Usingthe result of Problem 1.9, set the potential energy of this system equal to mc2

and see what you get for ro. One defect of the model is rather obvious: Nothingis provided to hold the charge together!

o dr

dq

rρ

r

Figure 1: Diagram for Problem 1.10.In order to determine the potentialenergy of a sphere, we add sphericalshells of thickness dr.

The spherical charge of radius ro and charge e has charge density

ρ =e

V=

3e

4πr3o

if we had a smaller sphere of radius r with the same charge density, the potential at thesurface of such a sphere would be

Φ =ρV (r)

r=

ρ4πr3

3r

=4πρr2

3

1

consider a spherical shell of thickness dr, brought in from infinity to the surface of thissphere. The work done must then be

dW = Φdq = Φ4πr2ρdr

=(4πρ)2

3r4dr

W =

∫ ro

o

(4πρ)2

3r4dr

=(4πρ)2

15r5dr

=3

5e2/ro

which is the same as the total potential energy of the field.Setting this quantity equal to mc2 and solving for ro we obtain

mc2 = U =3

5e2/ro

⇒ ro =3

5

e2

mc2

=3

5

(4.80 × 10−10esu)2

(0.91 × 10−27g)(3 × 1010cm/s)2

⇒ ro = 1.69 × 10−12 cm (1)

2

Purcell §1.30

Concentric spherical shells of radius a and b, with b > a, carry charge Q and−Q, repectively, each charge uniformly distributed. Find the energy stored inthe electric field of this system.

The electric field due to two concentric oppositely charged sphere can be found usingGauss’s law. Taking our Gaussian surface outside the outer sphere, or inside the inner

Gaussian Surface

ab

r

+Q−Q

Figure 2: Diagram for Problem 1.30.The electric field between two chargedconcentric spheres can be found usingGauss’s law.

sphere we quickly see that there can be no electric field in these regions, as there is no netcontained charge.

In the region between the two shells, the electric field must be radially directed, with flux

Φ = E · 4πr2 = 4πQ

⇒ E =Q

r2

Knowing the electric field everywhere in space we can then integrate to find the totalenergy contained in the electric field.

3

U =1

8π

∫

∞

E2dV

=1

8π

∫ ∞

0

E24πr2dr

=1

2

∫ b

a

Q2

r4r2dr

⇒ U =Q2

2

(

1

a− 1

b

)

4

Purcell §2.1

The vector which follows represents a possible electrostatic field:

Ex = 6xy Ey = 3x2 − 32y Ez = 0

Calculate the line integral of ~E from the point (0,0,0) to the point (x1,y1, 0) alongthe path which runs straight from (0,0,0) to (x1, 0, 0) and thence to (x1,y1, 0).Make a similar calculation for the path which runs along the other two sides ofthe rectangle, via the point (0, y1, 0). You ought to get the same answer if theassertion above is true. Now you have the potential function φ(x, y, z). Take thegradient of this function and see that you get back the components of the givenfield.

path 1

(0, y, 0)

(x, 0, 0)

(x, y, 0)

y

x

path 2

Figure 3: Diagram for Problem 2.1.The two paths taken yield the samevalue for the path integral. This indi-cates that the described field is conser-vative, and hence can be described bythe gradient of a potential.

Path 1:

−∇Φ =

∫ (x1,y1)

(0,0)

~E · d~s

=

∫ x1

0

Ex(x, 0)dx +

∫ y1

0

Ey(x1, y)dy

= 0 +

∫ y1

0

(3x21 − 3y2)dy

= 3x21y1 − y3

1 (2)

5

Path 2:

−∇Φ =

∫ (x1,y1)

(0,0)

~E · d~s

=

∫ y1

0

Ey(0, y)dy +

∫ x1

0

Ex(x, y1)dx

=

∫ y1

0

−3y2dy +

∫ x1

0

6xy1dx

= 3x21y1 − y3

1

Thus the electric potential, if taken as zero at (0,0) is 3y3 − 3x2y .Taking the gradientwe see that it returns the electric field.

Purcell §2.12

The right triangle with vertex P at the origin base b, and altitude a, has auniform density of surface charge σ. Determine the potential at the vertex P.First find the contribution of the vertical strip of width dx at x. Show that thepotential at P can be written as φP = σb ln[(1 + sin θ) cos θ].

This problem is more an exercise in integration than anything else.

xa/bdy

dx

a

b

Figure 4: Diagram for Problem 2.12.First we integrate the vertical slice lo-cated at position x. Then we integrateover x.

6

φP = σ

∫ b

0

dx

∫ xa

b

0

dy√

x2 + y2

= σ

∫ b

0

dx ln(y +√

x2 + y2)

∣

∣

∣

∣

xa

b

0

= σ

∫ b

0

dx ln(a

b+

√

1 +a2

b2)

= σb ln

(

a

b+

√

1 +a2

b2

)

⇒ φP = σb ln(1 + sin θ

cos θ) (3)

since ab

+√

a2+b2

b= sin θ+1

cos θ.

Purcell §2.18

A hollow circular cylinder, of radius a, and length b, with open ends, has atotal charge Q uniformly distributed over its surface. What is the difference inpotential between a point on the axis at one end and the midpoint of the eaxis?Show by sketching some field lines how you think the field of this thing ought tolook.

Since all charge on a ring is equidistant from a point on the axis, the ring is a convenientblock on which to construct our potential. Consider a point located a distane xo from themidpoint. (Right side of Fig[5]). The potential can be expressed as:

φ =

∫

dQ

r

=

∫ b

2+xo

− b

2+xo

(

Qdx

b

)

1√a2 + x2

=Q

b[ln(

√a2 + x2 + x)]

∣

∣

∣

∣

b

2+xo

− b

2+xo

For the midpoint we have xo = 0:

φA =Q

bln

(

√

a2 + b2/4 + b/2√

a2 + b2/4 − b/2

)

7

dQ = Qdx/b

a

b

dx

xo

b/2

ar

x

Figure 5: Diagram for Problem 2.18. The total charge Q is uniformly distributed.thus the charge in the ring of width dx is dQ = Q dx

b. All charge on the ring is

equidistant from the axis.

For a point at the end of the tube we have xo = b/2:

φB =Q

bln

(√

a2 + b2 + b

a

)

⇒ φA − φB =Q

bln

(

a(√

a2 + b2/4 + b/2)

(√

a2 + b2/4 − b/2)(√

a2 + b2 + b)

)

(4)

Figure 6: Sketch of the elelctric fields.

8

Purcell §2.27

Use the result stated in Eq. 24 to calculate the energy stored in the electric fieldof the charged disk described in Section 2.6. (Hint: Consider the work done inbuilding the disc of charge out from zero radius to radius a by adding successiverings of width dr. Express the total energy in terms of radius a and total chargeQ = πa2σ).

r

dr

Figure 7: Diagram for Problem 2.27. The potential at the rim of the chargedidsk is 4σr. We add a ring of charge dq = σ2πrdr.

The potential of at the rim of a charged disc derived in the book and in section is givenby

φ(r) = 4σr

where r is the radius of the disc. Adding a ring of charge dq = σ2πrdr costs in energy

dE = φdq

= 8πσ2r2dr

E =

∫ a

0

8πσ2r2dr

=8π

3σ2a3

But we have that σ = Q/(πa2) such that

⇒ E =8Q2

3πa(5)

9