Question

There is a spherical conductor of radius R centered at $C_1$ from which sphere of radius R/2 centered at $C_2$ is carved out. A point charge 2q is placed at distance R/4 in the cavity & charge $-q$ is placed on surface of conductor as shown. Electric potential at $C_2$ is

Answer 1

$\frac{7q}{4\pi\epsilon_0 R}$

Answer 2

The electric potential at any point within the conductor material is constant. Let this potential be $V_0$. The inner surface of the cavity is part of the conductor boundary, so the potential at any point on the inner surface is $V_0$. The outer surface of the conductor is also at potential $V_0$.

The potential at any point $P$ inside the cavity is the superposition of the potential due to the point charge $2q$ and the potential due to the induced charge on the inner surface of the cavity. Let $V_{2q}(P)$ be the potential due to the point charge $2q$ and $V_{induced}(P)$ be the potential due to the induced charge on the inner surface. So, for any point $A$ on the inner surface of the cavity, the potential is $V_0 = V_{2q}(A) + V_{induced}(A) + V_{outer}(A)$, where $V_{outer}(A)$ is the potential at $A$ due to the charge on the outer surface of the conductor. Since the outer surface is a sphere of radius $R$ centered at $C_1$, and the point $A$ is inside this sphere, the potential due to the outer surface charge is constant everywhere inside the outer sphere. Let this constant potential be $V_{outer}$.

Consider the potential at $C_2$. $C_2$ is inside the cavity. The potential at $C_2$ is $V(C_2) = V_{2q}(C_2) + V_{induced}(C_2) + V_{outer}(C_2)$. Since $V_{outer}$ is constant inside the outer sphere, $V_{outer}(C_2) = V_{outer}$.

Let's consider the potential difference between a point $A$ on the inner surface and $C_2$. $V(A) - V(C_2) = (V_{2q}(A) + V_{induced}(A) + V_{outer}) - (V_{2q}(C_2) + V_{induced}(C_2) + V_{outer})$ $V(A) - V(C_2) = (V_{2q}(A) + V_{induced}(A)) - (V_{2q}(C_2) + V_{induced}(C_2))$. Since $A$ is on the inner surface, $V(A) = V_0$. So, $V_0 - V(C_2) = (V_{2q}(A) + V_{induced}(A)) - (V_{2q}(C_2) + V_{induced}(C_2))$.

Consider the potential inside the cavity due to the point charge $2q$ and the induced charge on the inner surface. This potential is such that the potential on the inner surface is constant ($V_0 - V_{outer}$). Let $V'(P) = V_{2q}(P) + V_{induced}(P)$. Then $V'(A) = V_0 - V_{outer}$ for any point $A$ on the inner surface. The potential inside the cavity is determined by the point charge inside and the boundary condition on the inner surface. Consider the potential difference between $C_2$ and a point $A$ on the inner surface due to the point charge $2q$ and the induced charge. $V'(C_2) - V'(A) = V_{2q}(C_2) - V_{2q}(A) + V_{induced}(C_2) - V_{induced}(A)$. Since $V'(A) = V_0 - V_{outer}$, $V'(C_2) = V(C_2) - V_{outer}$. So, $V(C_2) - V_{outer} - (V_0 - V_{outer}) = V(C_2) - V_0 = V_{2q}(C_2) - V_{2q}(A) + V_{induced}(C_2) - V_{induced}(A)$.

A key property is that the potential at any point inside a conductor is constant. Also, the potential on the surface of a conductor is constant. The electric field inside the conductor is zero. Consider the potential at $C_2$. The potential at $C_2$ is the sum of the potential due to the point charge $2q$, the induced charge on the inner surface, and the charge on the outer surface. The potential inside the conductor is constant, say $V_0$. This constant potential extends up to the inner surface of the cavity.

Consider the potential at $C_2$ due to the point charge $2q$ and the induced charges on the inner surface. For a point charge $q$ inside a spherical cavity of radius $a$ in a conductor, the potential at any point $r$ inside the cavity is given by $V(r) = \frac{q}{4\pi\epsilon_0 |r-r_0|} + V_{induced}(r)$. The potential on the inner surface is constant. If the inner surface is at potential $V_{surf}$, then $V_{induced}(r) = V_{surf} - \frac{q}{4\pi\epsilon_0 |r-r_0|}$. The potential at any point $r$ inside the cavity is then $V(r) = \frac{q}{4\pi\epsilon_0 |r-r_0|} + V_{surf} - \frac{q}{4\pi\epsilon_0 |r-r_0|} = V_{surf}$. This is incorrect.

Let's consider the potential at $C_2$ due to the point charge $2q$ and the induced charge on the inner surface. The potential on the inner surface is $V_0$. The potential at $C_2$ is inside the cavity. The potential at $C_2$ due to the point charge $2q$ at distance $R/4$ is $\frac{1}{4\pi\epsilon_0} \frac{2q}{R/4} = \frac{8q}{4\pi\epsilon_0 R}$. The induced charge on the inner surface is $-2q$.

Consider the potential at the center of a charged spherical shell. It is the same as the potential on the surface. Consider the potential at $C_2$ due to the induced charge on the inner surface. The induced charge is distributed on the inner surface such that the potential on the inner surface due to the point charge $2q$ and the induced charge is constant. Let $V_{in}(r)$ be the potential inside the cavity due to the point charge and the induced charge. $V_{in}(r) = V_{2q}(r) + V_{induced}(r)$. On the inner surface (radius $R/2$ around $C_2$), $V_{in}(r) = V_0 - V_{outer}$. The potential at $C_2$ is $V(C_2) = V_{in}(C_2) + V_{outer}$.

Let's consider the potential at $C_2$ due to the point charge $2q$ and the induced charge. The potential on the inner surface is $V_i = V_0 - V_{outer}$. The potential at $C_2$ due to a point charge $q$ at distance $d$ from the center of a spherical shell of radius $a$ with charge $Q$ on it is $\frac{1}{4\pi\epsilon_0} (\frac{q}{d} + \frac{Q}{a})$. This is for a solid sphere.

Let's use the property that the potential is constant inside the conductor. Consider the potential at $C_1$. $C_1$ is inside the conductor. So $V(C_1) = V_0$. The potential at $C_1$ is the sum of potentials due to the point charge $2q$, the induced charge on the inner surface, and the charge $-q$ on the outer surface. The distance from the point charge to $C_1$ is the distance from $C_1$ to $C_2$ plus the distance from $C_2$ to the point charge, if they are along the same line. Assuming $|C_1 - C_2| = R/2$ and the point charge is at distance $R/4$ from $C_2$. If $C_1$, $C_2$, and the point charge are collinear, the distance from $C_1$ to the point charge is $R/2 \pm R/4$.

However, we know that the potential on the inner surface is constant, equal to $V_0$. The potential at $C_2$ is inside the cavity. The potential at $C_2$ due to the point charge $2q$ is $\frac{1}{4\pi\epsilon_0} \frac{2q}{R/4} = \frac{8q}{4\pi\epsilon_0 R}$. The potential at $C_2$ due to the induced charge on the inner surface and the charge on the outer surface is constant everywhere inside the conductor and in the cavity. This constant potential is equal to the potential of the conductor, $V_0$. This is because the region outside the cavity is a conductor, and the electric field due to charges outside a closed conducting surface is zero everywhere inside the surface. So, the induced charge on the inner surface and the charge on the outer surface together produce a constant potential inside the cavity. Let $V_{rest}(P)$ be the potential at point $P$ due to the induced charge on the inner surface and the charge on the outer surface. Then $V_{rest}(P) = V_0$ for all points $P$ inside the cavity. Therefore, the potential at $C_2$ is $V(C_2) = V_{2q}(C_2) + V_{rest}(C_2)$. $V(C_2) = \frac{1}{4\pi\epsilon_0} \frac{2q}{R/4} + V_0 = \frac{8q}{4\pi\epsilon_0 R} + V_0$.

Now we need to find $V_0$. The potential of the conductor is $V_0$. This potential is determined by the total charge on the conductor and the external charges. The total charge on the conductor is the induced charge on the inner surface ($-2q$) plus the charge on the outer surface ($-q$). So the total charge on the conductor is $-3q$. The potential on the outer surface of a conductor with total charge $Q_{total}$ and influenced by external charges is generally not simply $k Q_{total} / R$.

However, consider the potential at $C_2$. $V(C_2) = V_{2q}(C_2) + V_{induced}(C_2) + V_{-q}(C_2)$. The potential due to the induced charge on the inner surface and the charge on the outer surface is constant inside the cavity, and equal to the potential of the conductor $V_0$. So, $V_{induced}(C_2) + V_{-q}(C_2) = V_0$. Thus, $V(C_2) = V_{2q}(C_2) + V_0$. $V_{2q}(C_2) = \frac{1}{4\pi\epsilon_0} \frac{2q}{R/4} = \frac{8q}{4\pi\epsilon_0 R}$.

To find $V_0$, consider the potential at the outer surface. The outer surface is at potential $V_0$. The potential at a point on the outer surface is the sum of potentials due to the point charge $2q$, the induced charge on the inner surface, and the charge $-q$ on the outer surface. The potential at a point on the outer surface due to the point charge $2q$ and the induced charge on the inner surface is constant everywhere outside the inner surface. This is because the total charge inside the inner surface is $2q + (-2q) = 0$. So, the potential outside the cavity due to the charges inside the cavity is zero. Therefore, the potential at any point on the outer surface is solely due to the charge $-q$ on the outer surface. For a spherical shell of radius $R$ with charge $-q$ uniformly distributed on its surface, the potential on the surface is $\frac{1}{4\pi\epsilon_0} \frac{-q}{R}$. So, $V_0 = \frac{1}{4\pi\epsilon_0} \frac{-q}{R}$.

Substituting this value of $V_0$ into the expression for $V(C_2)$: $V(C_2) = \frac{8q}{4\pi\epsilon_0 R} + \frac{-q}{4\pi\epsilon_0 R} = \frac{7q}{4\pi\epsilon_0 R}$.