Multiple Choice Questions

1. A parallel-plate capacitor with a dielectric material between its plates is charged to a potential difference V. The dielectric is then removed. What happens to the potential difference between the plates?

   (A) It increases. (B) It decreases. (C) It remains the same. (D) It becomes zero.

2. A capacitor is charged and then disconnected from the battery. A dielectric material is inserted between the plates. Which of the following quantities will decrease?

   (A) Charge on the capacitor (B) Capacitance of the capacitor (C) Potential difference across the capacitor (D) Energy stored in the capacitor

Free Response Question

A capacitor consists of two conducting, coaxial, cylindrical shells of radius a and b, respectively, and length L >> b. The space between the cylinder is filled with oil that has a dielectric constant k. Initially both cylinders are uncharged, but then a battery is used to charge the capacitor, leaving a charge +Q on the inner cylinder and -Q on the outer cylinder, as shown above. Let r be the radial distance from the axis of the capacitor.

(a) Using Gauss's Law, determine the electric field midway along the length of the cylinder for the following values of r, in terms of the given quantities and fundamental constants. Assume end effects are negligible.

i. *a* < *r* < *b*
ii. *b* < *r* << *L*

(b) Determine the following in terms of the given quantities and fundamental constants.

i. The potential difference across the capacitor.
ii. The capacitance of this capacitor.

Answer Key

Multiple Choice Answers

1. (A) It increases.
2. (C) Potential difference across the capacitor

Free Response Answers

(a)

• i. a < r < b

  • Apply Gauss's Law:

    $$\oint \vec{E} \cdot d\vec{A} = \frac{Q_{enc}}{\epsilon_0}$$

  • The Gaussian surface is a cylinder of radius r and length L.

  • The electric field is radial and constant over the Gaussian surface:

    $$E(2\pi r L) = \frac{Q}{\epsilon_0 k}$$

    $$E = \frac{Q}{2\pi \epsilon_0 k r L}$$

    • 1 point for correct application of Gauss's Law with dielectric
    • 1 point for correct electric field expression

• ii. b < r << L

  • The Gaussian surface now encloses both the inner and outer cylinders, which have equal and opposite charges (+Q and -Q).

  • $Q_{enc} = +Q - Q = 0$

    $$E = 0$$

    • 1 point for recognizing that the enclosed charge is zero
    • 1 point for correct electric field expression

(b)

• i. Potential Difference

  • Use the integral form of potential difference:

    $$V = -\int_a^b \vec{E} \cdot d\vec{r}$$

  • Since the electric field is radial and the path is along the radial direction, the dot product becomes a simple product.

    $$V = -\int_a^b \frac{Q}{2\pi \epsilon_0 k r L} dr$$

    $$V = -\frac{Q}{2\pi \epsilon_0 k L} \int_a^b \frac{1}{r} dr$$

    $$V = -\frac{Q}{2\pi \epsilon_0 k L} [\ln(r)]_a^b$$

    $$V = \frac{Q}{2\pi \epsilon_0 k L} \ln(\frac{a}{b})$$

    • 1 point for correct integral setup
    • 1 point for correct integration
    • 1 point for correct potential difference expression

• ii. Capacitance

  • Use the definition of capacitance: $C = \frac{Q}{V}$

    $$C = \frac{Q}{\frac{Q}{2\pi \epsilon_0 k L} \ln(\frac{a}{b})}$$

    $$C = \frac{2\pi \epsilon_0 k L}{\ln(\frac{b}{a})}$$

    • 1 point for correct expression of capacitance