Calculation of the Electric Potential of a Uniformly Charged Spherical Shell

In the study of electrostatics, systems exhibiting spherical symmetry represent some of the most fundamental and instructive models. Among these, the uniformly charged spherical shell serves as a cornerstone for understanding the relationship between electric field intensity and electric potential. It provides a clear, mathematical demonstration of how Gauss's Law dictates field behavior and how the continuity of scalar fields differs from the behavior of vector fields.

Consider a thin spherical shell of radius $R$, upon which a total charge $Q$ is distributed uniformly. The surface charge density, $\sigma$, is defined as:
$$\sigma = \frac{Q}{4\pi R^2}$$

Our objective is to determine the electric potential $\Phi(r)$ at any point $P$ in space, where $r$ represents the distance from the center of the shell. By convention, we define the potential at infinity to be zero: $\Phi(\infty) = 0$.
Before calculating the potential, we must first establish the electric field $\mathbf{E}$ produced by the shell. Due to the spherical symmetry of the charge distribution, the electric field must be purely radial, and its magnitude $E$ can only depend on the radial distance $r$. We apply Gauss's Law to two distinct regions:

1. Outside the Shell ($r > R$)

To find the field external to the shell, we imagine a spherical Gaussian surface with radius $r > R$. According to Gauss's Law, the electric flux through this surface is proportional to the enclosed charge:
$$\oint \mathbf{E} \cdot d\mathbf{A} = \frac{Q_{encl}}{\epsilon_0} \implies E \cdot 4\pi r^2 = \frac{Q}{\epsilon_0}$$
Solving for $E$, we get:
$$E_{out} = \frac{Q}{4\pi\epsilon_0 r^2} \quad (r > R)$$
This result demonstrates that, from an external perspective, the spherical shell behaves exactly like a point charge located at the center.

2. Inside the Shell ($r < R$)

For a Gaussian surface with radius $r < R$, the surface resides entirely within the hollow interior of the shell. Since the charge $Q$ is distributed strictly on the surface, the net enclosed charge $Q_{encl}$ is zero.
$$E \cdot 4\pi r^2 = 0 \implies E_{in} = 0$$
Consequently, the electric field inside a uniformly charged spherical shell is zero at all points.

Deriving the Electric Potential Distribution

The electric potential $\Phi(r)$ is related to the electric field $\mathbf{E}$ through the line integral:
$$\Phi(r) = -\int_{\infty}^{r} \mathbf{E} \cdot d\mathbf{l}$$

Because the electric field follows different functional forms depending on the position relative to $R$, we must evaluate the potential in two separate cases.

1. Potential in the External Region ($r \ge R$)

For any point outside or on the surface of the shell, we integrate the external field from infinity to the distance $r$:
$$\Phi(r) = -\int_{\infty}^{r} \frac{Q}{4\pi\epsilon_0 r'^2} dr'$$
$$\Phi(r) = \left[ \frac{Q}{4\pi\epsilon_0 r'} \right]_{\infty}^{r} = \frac{Q}{4\pi\epsilon_0 r} - 0$$
Thus, for $r \ge R$:
$$\Phi(r) = \frac{Q}{4\pi\epsilon_0 r}$$

2. Potential in the Internal Region ($r < R$)

Calculating the potential inside the shell requires a two-step integration path. We must first integrate from infinity to the shell's surface ($R$), and then from the surface to the internal point ($r$):
$$\Phi(r) = -\int_{\infty}^{R} \mathbf{E}{out} \cdot d\mathbf{l} - \int{R}^{r} \mathbf{E}_{in} \cdot d\mathbf{l}$$

We already know that the potential at the surface ($r=R$) is $\Phi(R) = \frac{Q}{4\pi\epsilon_0 R}$. Since the internal electric field $\mathbf{E}{in}$ is zero, the second integral vanishes:
$$\Phi(r) = \Phi(R) - \int{R}^{r} 0 \cdot dr' = \Phi(R)$$
Therefore, for $r < R$:
$$\Phi(r) = \frac{Q}{4\pi\epsilon_0 R}$$

Summary and Physical Analysis

The complete expression for the electric potential of a uniformly charged spherical shell is:
$$\Phi(r) =
\begin{cases}
\frac{Q}{4\pi\epsilon_0 r}, & r \ge R \
\frac{Q}{4\pi\epsilon_0 R}, & r < R
\end{cases}$$

Key Physical Insights

• Continuity of Potential: As $r$ approaches $R$ from either the outside or the inside, the value of $\Phi(r)$ converges to $\frac{Q}{4\pi\epsilon_0 R}$. This confirms that the electric potential is a continuous function. In physics, a sudden jump in potential would imply an infinite amount of energy, which is non-physical for a finite charge distribution.
• Discontinuity of the Electric Field: While the potential is continuous, its gradient (the electric field) is not. There is a sharp jump in the field magnitude at $r=R$, transitioning from $\frac{Q}{4\pi\epsilon_0 R^2}$ to zero. This discontinuity is a direct consequence of the presence of the surface charge density $\sigma$.
• The Equipotential Interior: Inside the shell, the potential is constant and independent of $r$. This means the entire interior volume is an equipotential region. Because the electric field is the negative gradient of the potential ($\mathbf{E} = -\nabla \Phi$), a constant potential naturally results in a zero electric field.

Practical Application: Calculating Electrostatic Work

To illustrate the utility of these derivations, consider the following scenario:

Problem: A test charge $q$ is moved from infinity to a point $P$ located anywhere inside the shell ($r < R$). How much work must an external agent perform to move this charge?

Solution:

1. Relate Work to Potential Energy: The work done by an external agent ($W_{ext}$) is equal to the change in the electrostatic potential energy ($U$) of the system:
   $$W_{ext} = \Delta U = U(r) - U(\infty)$$
2. Apply the Potential Formula: The potential energy of a charge $q$ at a point $r$ is $U(r) = q \cdot \Phi(r)$. Since we defined $\Phi(\infty) = 0$, the initial potential energy is zero.
3. Calculate the Result: Since the destination point $r$ is inside the shell, we use the internal potential $\Phi(r) = \frac{Q}{4\pi\epsilon_0 R}$:
   $$W_{ext} = q \left( \frac{Q}{4\pi\epsilon_0 R} \right) = \frac{qQ}{4\pi\epsilon_0 R}$$

Conclusion: Interestingly, the work required to move the charge depends only on the shell's radius $R$ and the total charge $Q$, not on the specific final distance $r$, provided the charge ends up anywhere inside the shell. Once the charge is inside, no further work is required to move it between any two internal points because the internal electric field is zero.