Physics 7D Makeup Midterm Solutions

Problem 1 A hollow, conducting sphere with an outer radius of 0.253 m and an inner radius of 0.194 m has a uniform surface charge density of 6.96 × 10−6 C/m2. A charge of −0.510 µC is now introduced into the cavity inside the sphere. (a) What is the new charge density on the outside of the sphere? (b) Calculate the strength of the electric field just outside the sphere. (c) What is the electric flux through a spherical surface just inside the inner surface of the sphere?

Solution (a) Because there is no electric field inside a conductor the sphere must acquire a charge of +0.510 µC on its inner surface. Because there is no net flow of charge onto or off of the conductor this charge must come from the outer surface, so that the total charge of the outer 2 surface is now Qi − 0.510µC = σ(4πrout) − 0.510µC = 5.09µC. Now we divide again by the surface area to obtain the new charge density.

5.09 × 10−6C σ = = 6.32 × 10−6C/m2 (1) new 4π(0.253m)2

(b) The electric field just outside the sphere can be determined by Gauss’ law once we determine the total charge on the sphere. We’ve already accounted for both the initial surface charge and the newly introduced point charge in our expression above, which tells us that −6 Qenc = 5.09 × 10 C. Gauss’ law then tells us that

Qenc 5 E = 2 = 7.15 × 10 N/C (2) 4π0rout (c) Finally we want to find the electric flux through a spherical surface just inside the sphere’s cavity. From Gauss’ law we have that the electric flux is just Qenc/0. In this case the enclosed charge is just the new charge −0.510µC, so our electric flux is

−0.510µC 4 2 ΦE = = −5.76 × 10 Nm /C (3) 0

Problem 2

A point charge q1 = 2.0µC is located at (−0.11m, −0.06m), a second point charge q2 = −5.0µC is located at (0.15m, 0.11m), and a third point charge q3 = 3.0µC is located at (0.7m, −0.30m). Calculate the work required to assemble such a system. 1 Solution

To assemble this system we start with the first point charge q1. In the absence of any external potential it takes no work to place this charge anywhere. Next we bring in the second point charge from infinity. This takes negative work because the second point charge is attracted to the first. The potential energy in question is

kq1q2 k(2.0µC)(−5.0µC) U2 = = p = −0.289 J (4) r12 (0.15 + 0.11)2 + (0.11 + 0.06)2 m

Now that our first two charges are situated we bring in the third charge. It has two contributions to its energy: the potential energy caused by the interaction between it and the first charge and that due to the second charge

kq1q3 kq2q3 U3 = + r13 r23 k(2.0µC)(3.0µC) k(−5.0µC)(3.0µC) = + (5) p(0.7 + 0.11)2 + (−0.3 + 0.06)2 m p(0.7 − 0.15)2 + (−0.30 − 0.11)2 m = −0.133 J

Therefore the work that was done to assemble this configuration is

W = 0.422 J (6)

Problem 3

The disk with a circular hole at its center (an annulus) has inner radius R1 and outer radius R2. The disk has a uniform surface charge density σ. The annulus lies in the yz−plane with its center at the origin. Determine the electric field for an arbitrary point on the x-axis.

Solution To solve this problem we use superposition. First I parameterize the annulus: any point lying on the annulus is completely described by its distance r from the x-axis and an angle θ from some arbitrary starting point. A differential piece of the annulus has charge dq = σr dr dθ, and we can write the electric field magnitude due to this piece of the annulus as

σr dr dθ 2 2 (7) 4π0 (r + x )

2 By symmetry considerations the z and y components of the electric field will cancel, so we need only consider the x component. We extract this from the magnitude given above by multiplying by a trigonometric factor.

σr dr dθ x √ dEx = 2 2 (8) 4π0 (r + x ) x2 + r2 Now we evaluate the total electric field by integrating over the annulus

Z R2 Z 2π σrx Ex = dr dθ 2 2 3/2 R1 0 4π0(r + x ) σx Z R2 rdr = 2 2 3/2 20 R1 (r + x ) R (9) σx  1  2 = − 2 (r2 + x2)1/2 0 R1 " # σx 1 1 = − 2 p 2 2 p 2 2 0 R1 + x R2 + x

Problem 4 A cylindrical air capacitor with a length of 17.0 m stores an amount of energy equal to 3.60×10−9 J when the potential difference between the two conductors is 4.80 V. (a) Calculate the magnitude of the charge on each conductor. (b) Calculate the ratio of the radii of the inner and outer conductors.

Solution (a) For this part we don’t need to know anything about the capacitor. We start with U = Q2/2C = QV/2. Then solving for Q we have

2U 7.20 × 10−9 J Q = = = 1.5 nC (10) V 4.80 V (b) Now we do need to know something about cylindrical capacitors. If we place a charge of Q on the inner cylinder then the electric field in the space between the two conductors will be (ignoring fringing effects and taking the dielectric constant of air to be negligible)

Q/` E = (11) 2π0r

3 where r is the distance from the cylinder’s axis and ` is the length of the capacitor. The potential difference will then be

Z b Q b |∆V | = Edr = log (12) a 2π0` a where a and b are the radii of the cylinders, so that the capacitance is

Q 2π ` C = = 0 (13) ∆V log(b/a)

Now we use our other expression for the energy

CV 2 2U U = ⇒ C = (14) 2 V 2 and we can finally solve for the ratio of the radii.

b V 2π l (4.80V)2π (17.0m) = exp 0 = exp 0 = 20.6 (15) a U 3.6 × 10−9J

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