AP Physics C- Electricity and Magnetism- 9.2 Electric Potential- Exam Style questions - FRQs- New Syllabus

Here is a detailed explanation of the problem. We start by using properties of conductors in electrostatic equilibrium to find how charge distributions spread out over a metal shell. From there, Gauss’s law lets us look inside the insulating sphere by calculating exactly how much charge is enclosed inside our surface relative to its total volume. Finally, relating the electric field back to potential difference involves integrating over the radial path between the boundaries of our system.

(a)
Inside the conducting shell, the electric field must be zero in electrostatic equilibrium:
\(E = 0\)
Therefore, the net charge enclosed by a Gaussian surface inside the conducting material must be zero:
\(q_{\text{enc}} = 0\)
\(q_{\text{inner\_surface}} + q_{\text{sphere}} = 0\)
\(q_{\text{inner\_surface}} + (-Q) = 0\)
\(q_{\text{inner\_surface}} = +Q\)
The net charge on the shell is the sum of the charges on its inner and outer surfaces:
\(q_{\text{net}} = q_{\text{inner\_surface}} + q_{\text{outer\_surface}}\)
\(+3Q = +Q + q_{\text{outer\_surface}}\)
\(q_{\text{outer\_surface}} = +2Q\)

(b)
Apply Gauss’s law using a concentric spherical surface of radius \(r < R\):
\(\oint E \cdot dA = \dfrac{q_{\text{enc}}}{\varepsilon_0}\)
\(E(4\pi r^2) = \dfrac{q_{\text{enc}}}{\varepsilon_0}\)
The uniform volume charge density is:
\(\rho = \dfrac{-Q}{\frac{4}{3}\pi R^3}\)
The charge enclosed within radius \(r\) is:
\(q_{\text{enc}} = \rho V_{\text{enc}} = \left(\dfrac{-Q}{\frac{4}{3}\pi R^3}\right)\left(\dfrac{4}{3}\pi r^3\right) = -Q\dfrac{r^3}{R^3}\)
Substitute \(q_{\text{enc}}\) back into the Gauss’s law expression:
\(E(4\pi r^2) = \dfrac{-Q r^3}{\varepsilon_0 R^3}\)
\(E = -\dfrac{Qr}{4\pi\varepsilon_0 R^3}\)

(c)
For \(r \ge R\), the electric field outside the nonconducting sphere behaves like that of a point charge enclosed at the center:
\(E \propto \dfrac{1}{r^2}\)
At \(r = R\), the field magnitude is given as:
\(E_1 = 8\,\text{N/C}\)
At \(r = 2R\), the distance doubles, so the field breaks down by a factor of four:
\(E_2 = \dfrac{E_1}{2^2} = \dfrac{8}{4} = 2\,\text{N/C}\)
\(\boxed{E = 2\,\text{N/C}}\)

(d)
The potential difference between the outer surface of the sphere (\(r = R\)) and the inner surface of the shell (\(r = 4R\)) is found by integrating the electric field in that region:
\(\Delta V = -\int_{R}^{4R} E \cdot dr\)
In the region \(R < r < 4R\), the field is due solely to the inner sphere’s net charge \(-Q\):
\(E = -\dfrac{Q}{4\pi\varepsilon_0 r^2}\)
Set up the absolute potential integral:
\(|\Delta V| = \left| -\int_{R}^{4R} \left(-\dfrac{Q}{4\pi\varepsilon_0 r^2}\right) dr \right|\)
\(|\Delta V| = \dfrac{Q}{4\pi\varepsilon_0} \left[ -\dfrac{1}{r} \right]_{R}^{4R}\)
\(|\Delta V| = \dfrac{Q}{4\pi\varepsilon_0} \left( \dfrac{1}{R} – \dfrac{1}{4R} \right)\)
\(|\Delta V| = \dfrac{3Q}{16\pi\varepsilon_0 R}\)

(e)(i)

• In region I (\(0 < r < R\)), \(E\) is negative and its magnitude increases linearly from zero to \(R\).
• In region II (\(R < r < 4R\)), \(E\) remains negative, starting at \(-8\,\text{N/C}\) at \(r = R\) and asymptotically approaching zero according to \(\dfrac{1}{r^2}\).
• In region III (\(r > 4R\)), the total enclosed net charge is \((-Q) + (+3Q) = +2Q\), so the field steps up to a positive value and decreases as a concave-up curve toward zero.

(e)(ii)

• From \(0\) to \(4R\), the electric potential curve is continuous and always increasing because the electric field points opposite to the direction of increasing \(r\).
• For \(r > 4R\), since the net outer charge is positive, the potential is positive and decreases asymptotically toward zero as \(r\) goes to infinity.