AP Physics C- Electricity and Magnetism- 10.3 Capacitors - Exam Style questions - FRQs- New Syllabus

Question

An isolated, air-filled, charged capacitor consists of two conducting, coaxial, cylindrical shells that each have length \(L\). The inner shell has radius \(R_1\) and the outer shell has radius \(R_2\), as shown in Figure \(1\), where \(R_1<R_2<L\). The surface charge densities, amounts of charge per unit area, on the inner and outer shells are \(+\sigma_1\) and \(-\sigma_2\), respectively. The absolute values of the total charges on the shells are equal.

i. Using Gauss’s law, derive an expression for the magnitude \(E\) of the electric field as a function of the radial distance \(r\) from the center of the capacitor for \(R_1<r<R_2\). Express your answer in terms of \(R_1\), \(\sigma_1\), \(r\), and physical constants, as appropriate.

ii. Derive an expression for the absolute value \(|\Delta V|\) of the potential difference between the outer and inner shells in terms of \(R_1\), \(R_2\), \(\sigma_1\), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

iii. On the axes shown in Figure \(2\), sketch a graph of \(E\) as a function of \(r\) from \(r=0\) to a position that is outside the outer shell.

B. A material of dielectric constant \(\kappa\) is inserted into the isolated, charged capacitor such that the material fills the region \(R_1<r<R_2\), as shown in Figure \(3\).

Derive an expression for the capacitance \(C\) of the capacitor with the material inserted in terms of \(L\), \(R_1\), \(R_2\), \(\kappa\), and physical constants, as appropriate. Begin your derivation by writing a fundamental physics principle or an equation from the reference information.

Most-appropriate topic codes (AP Physics C: Electricity and Magnetism):

• Topic \(8.4\) — Electric Fields of Charge Distributions (Part \( \mathrm{A(i)} \), Part \( \mathrm{A(iii)} \))
• Topic \(8.6\) — Gauss’s Law (Part \( \mathrm{A(i)} \), Part \( \mathrm{A(iii)} \))
• Topic \(9.2\) — Electric Potential (Part \( \mathrm{A(ii)} \))
• Topic \(10.1\) — Electrostatics with Conductors (Part \( \mathrm{A(iii)} \))
• Topic \(10.3\) — Capacitors (Part \( \mathrm{B} \))
• Topic \(10.4\) — Dielectrics (Part \( \mathrm{B} \))

▶️ Answer/Explanation

A(i)
Start with Gauss’s law:

\(\displaystyle \oint \vec{E}\cdot d\vec{A}=\dfrac{q_{\mathrm{enc}}}{\varepsilon_0}\)

Choose a cylindrical Gaussian surface of radius \(r\) and length \(\ell\), where \(R_1<r<R_2\). By symmetry, \(E\) is constant over the curved surface and is radial.

The flux through the curved surface is

\(\displaystyle \oint \vec{E}\cdot d\vec{A}=E(2\pi r\ell)\)

The charge enclosed is the charge on the portion of the inner cylindrical shell of length \(\ell\):

\(\displaystyle q_{\mathrm{enc}}=\sigma_1(2\pi R_1\ell)\)

Substitute into Gauss’s law:

\(\displaystyle E(2\pi r\ell)=\dfrac{\sigma_1(2\pi R_1\ell)}{\varepsilon_0}\)

Solving for \(E\),

\(\displaystyle \boxed{E=\dfrac{\sigma_1R_1}{\varepsilon_0 r}}\)

The electric field points radially outward from the positively charged inner shell.

A(ii)
The potential difference is related to the electric field by

\(\displaystyle \Delta V=-\int \vec{E}\cdot d\vec{r}\)

The magnitude of the potential difference between the shells is

\(\displaystyle |\Delta V|=\int_{R_1}^{R_2}E\,dr\)

Substitute \(E=\dfrac{\sigma_1R_1}{\varepsilon_0 r}\):

\(\displaystyle |\Delta V|=\int_{R_1}^{R_2}\dfrac{\sigma_1R_1}{\varepsilon_0 r}\,dr\)

\(\displaystyle |\Delta V|=\dfrac{\sigma_1R_1}{\varepsilon_0}\int_{R_1}^{R_2}\dfrac{1}{r}\,dr\)

\(\displaystyle |\Delta V|=\dfrac{\sigma_1R_1}{\varepsilon_0}\left[\ln r\right]_{R_1}^{R_2}\)

\(\displaystyle \boxed{|\Delta V|=\dfrac{\sigma_1R_1}{\varepsilon_0}\ln\!\left(\dfrac{R_2}{R_1}\right)}\)

A(iii)
The graph of \(E\) as a function of \(r\) has three main regions.

For \(0<r<R_1\), the electric field inside the conducting inner shell is zero:

\(\displaystyle E=0\)

For \(R_1<r<R_2\), the electric field decreases as \(\dfrac{1}{r}\):

\(\displaystyle E=\dfrac{\sigma_1R_1}{\varepsilon_0 r}\)

For \(r>R_2\), the total charge enclosed is zero because the two shells have equal and opposite total charges, so

\(\displaystyle E=0\)

Therefore, the graph should be zero before \(R_1\), then a decreasing concave-up curve between \(R_1\) and \(R_2\), and then zero again after \(R_2\).

B
Start with the definition of capacitance:

\(\displaystyle C=\dfrac{Q}{|\Delta V|}\)

From part \( \mathrm{A(ii)} \), without dielectric material,

\(\displaystyle |\Delta V|=\dfrac{\sigma_1R_1}{\varepsilon_0}\ln\!\left(\dfrac{R_2}{R_1}\right)\)

The charge on the inner shell is

\(\displaystyle Q=\sigma_1(2\pi R_1L)\)

Thus, the capacitance without the dielectric is

\(\displaystyle C_0=\dfrac{Q}{|\Delta V|}\)

\(\displaystyle C_0=\dfrac{\sigma_1(2\pi R_1L)}{\dfrac{\sigma_1R_1}{\varepsilon_0}\ln\!\left(\dfrac{R_2}{R_1}\right)}\)

Canceling \(\sigma_1\) and \(R_1\),

\(\displaystyle C_0=\dfrac{2\pi\varepsilon_0L}{\ln\!\left(\dfrac{R_2}{R_1}\right)}\)

When a dielectric of constant \(\kappa\) completely fills the region between the conducting shells, the capacitance increases by a factor of \(\kappa\):

\(\displaystyle C=\kappa C_0\)

Therefore,

\(\displaystyle \boxed{C=\dfrac{2\pi\kappa\varepsilon_0L}{\ln\!\left(\dfrac{R_2}{R_1}\right)}}\)

The dielectric reduces the electric field and potential difference for the same stored charge, so the ratio \(C=\dfrac{Q}{|\Delta V|}\) increases.

AP Physics C- Electricity and Magnetism- 10.3 Capacitors - Exam Style questions - FRQs- New Syllabus

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