AP® Physics C: E&M - Redistribution of Charge between Conductors - FRQs
Question
A capacitor consisting of conducting coaxial cylinders of radii a and b, respectively, and length L
is connected to a source of emf, as shown above. When the capacitor is charged, the inner cylinder
has a charge + Q on it. Neglect end effects and assume that the region between the cylinders is filled
with air. Express your answers in terms of the given quantities.
a. Use Gauss’s law to determine an expression for the electric field at a distance r from the axis of the
cylinder where a < r < b.
b. Determine the potential difference between the cylinders.
c. Determine the capacitance \(C_0\) of the capacitor.
One third of the length of the capacitor is then filled with a dielectric of dielectric constant k = 2, as
shown in the following diagram.
d. Determine the new capacitance C in terms of \(C_o\).
Answer/Explanation
Ans:
a. \(\oint E. dA = \frac{Q_{enc}}{\varepsilon_0}\)
\(E(2 \pi r L)=Q_{enc}/ \varepsilon_0 = Q/\varepsilon_0\)
\(E=Q/2 \pi \varepsilon_o L r\)
b. i.
\(\Delta V = -\int E. dr\)
\(V_C-V_D = -\int_{D}^{C} E. dr = \int_{C}^{D} E. dr = \int_{C}^{D} \frac{Q}{2 \pi \varepsilon_o L}\frac{dr}{r}=\frac{Q}{2 \pi \varepsilon_0 L} In r |^b_a = \frac{Q}{2 \pi \varepsilon_0 L} In \frac{b}{a}\)
c. \(C_0 =Q/V = 2 \pi varepsilon_0 L/In(b/a)\)
d. Consider the system as two capacitors in parallel, \(C=C_1 + C_2\)
\(C_{left}=kC_0/3\) and \(C_{right}=2/3C_0\) giving \(C=4/3 C_0\)
Question
Two concentric, conducting spherical shells, A and B, have radii a and b, respectively, (a < b).
Shell B is grounded, whereas shell A is maintained at a positive potential \(V_0\)
a. Using Gauss’s law, develop an expression for the magnitude E of the electric field at a distance r from
the center of the shells in the region between the shells. Express your answer in terms of the charge Q
on the inner shell.
b. By evaluating an appropriate integral, develop an expression for the potential Vo
c. Develop an expression for the capacitance of the system in terms of a and b
Answer/Explanation
Ans:
a. \(\oint E. dA = \frac{Q_{enc}}{\varepsilon_0}\)
\(E{4\pi r^2) = Q/\varepsilon_0\)
\(E=Q/4 \pi \varepsilon_0 r^2\)
b. \(\Delta V = kQ/a – kQ/b = kQ/b = kQ(b-a)/ab
c. C = Q/ \Delta V giving \(C=4 \pi \varepsilon_0 (ab/b – a)\)