A series of 3 uncharged concentric spherical conducting shells surround a small central charge q. The potential at a point outside the third shell, at distance r from the center, and relative to V = 0 at ∞, is

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Using Gauss's law, we know that the electric field at any point outside the charge q is the same as the electric field due to a point charge q located at the center of the shells. That is, $E = \frac{1}{4\pi\epsilon_0}\frac{q}{r^2}$ The electric potential at a point at a distance r from the center of the shells is given by integrating the electric field from ∞ to r. That is, $V = \int_\infty^r E \, dr = \int_\infty^r \frac{1}{4\pi\epsilon_0}\frac{q}{r^2} \, dr$ $V = \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r}\right)$ The potential due to each of the shells can be calculated separately by considering each shell as a point charge at its center. The potential due to a point charge located at a distance R from a point P is given by: $V = \frac{1}{4\pi\epsilon_0}\frac{q}{R}$ where q is the charge of the point charge. For each shell, the charge is the same and equal to -q (since each shell is uncharged). The potential due to the innermost shell is: $V_1 = \frac{1}{4\pi\epsilon_0}\frac{-q}{a}$ where a is the radius of the innermost shell. The potential due to the second shell is: $V_2 = \frac{1}{4\pi\epsilon_0}\frac{-q}{b}$ where b is the radius of the second shell. The potential due to the third shell is: $V_3 = \frac{1}{4\pi\epsilon_0}\frac{-q}{c}$ where c is the radius of the third shell. The total potential at a point outside the third shell is the sum of the potentials due to each of the shells and the potential due to the central charge q. That is, $V_{total} = V_1 + V_2 + V_3 + \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r}\right)$ Substituting the values for the radii of the shells, we get: $V_{total} = \frac{1}{4\pi\epsilon_0}\frac{-q}{a} + \frac{1}{4\pi\epsilon_0}\frac{-q}{b} + \frac{1}{4\pi\epsilon_0}\frac{-q}{c} + \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r}\right)$ $V_{total} = \frac{-q}{4\pi\epsilon_0}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right) + \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r}\right)$ $V_{total} = \frac{-q}{4\pi\epsilon_0}\left(\frac{bc+ac+ab}{abc}\right) + \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r}\right)$ $V_{total} = \frac{-q}{4\pi\epsilon_0}\left(\frac{ab+ac+bc}{abc}\right) + \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r}\right)$ $V_{total} = \frac{-q}{4\pi\epsilon_0}\left(\frac{a+b+c}{abc}\right) + \frac{q}{4\pi\epsilon_0}\left(\frac{1}{r}\right)$ Since the shells are concentric, we have: $a < b < c < r$ This implies that: $\frac{1}{a} > \frac{1}{b} > \frac{1}{c} > \frac{1}{r}$ Therefore, $\frac{a+b+c}{abc} > \frac{1}{abc}$ This means that the total potential is negative (since the first term is negative and the second term is positive). The correct answer is choice C.

A Series of 3 Uncharged Concentric Spherical Conducting Shells Surround