The problem presents two concentric spherical shells with radii $R_1$ and $R_2$, where $R_2 > R_1$. The shells carry charges $q_1$ and $q_2$, respectively. The task is to determine the electric potential in different regions:

1. $r < R_1$
2. $R_1 < r < R_2$
3. $r > R_2$

Solution

To find the electric potential $V(r)$ in each region, we apply Gauss's law, noting that:

• Inside a spherical shell with radius $R$ carrying charge $q$, the electric field is zero for $r < R$.
• Outside the shell, the field behaves as if all the charge were concentrated at the center.

The electric potential is calculated by integrating the electric field $E = \frac{1}{4 \pi \epsilon_0} \frac{Q}{r^2}$.

Let's break down the solution by region:

1. For $r < R_1$

In this region, we are inside the inner shell. Since there are no charges at $r < R_1$, the electric field is zero, and hence the potential is constant. We'll denote this constant potential as $V_1$, which we can determine based on continuity with the potential at $R_1$.

2. For $R_1 < r < R_2$

In this region, we are inside the outer shell but outside the inner shell. The electric field is due only to the charge $q_1$ on the inner shell. Using Gauss's law:

$E = \frac{1}{4 \pi \epsilon_0} \frac{q_1}{r^2}$

The potential at a point $r$ in this region can be found by integrating $E$ from $R_1$ to $r$:

$V(r) = V_1 - \int_{R_1}^{r} E \, dr = V_1 - \int_{R_1}^{r} \frac{1}{4 \pi \epsilon_0} \frac{q_1}{r^2} \, dr$

3. For $r > R_2$

In this region, the electric field is due to the combined charges $q_1 + q_2$. Using Gauss's law for $r > R_2$:

$E = \frac{1}{4 \pi \epsilon_0} \frac{q_1 + q_2}{r^2}$

The potential at a point $r > R_2$ can be found by integrating from $R_2$ to $r$:

$V(r) = V_2 - \int_{R_2}^{r} \frac{1}{4 \pi \epsilon_0} \frac{q_1 + q_2}{r^2} \, dr$

where $V_2$ is the potential at $R_2$, which can be calculated by summing up the potentials from the contributions of both shells.

Would you like further details or step-by-step calculations?

Related Questions

1. How does the electric potential vary inside a spherical shell?
2. What is the impact of a negative charge on one of the shells on the potential distribution?
3. How would the problem change if the shells had a dielectric material between them?
4. How can Gauss’s law be used to simplify electric field calculations for spherical symmetry?
5. How does electric field differ from electric potential in regions with no charge?

Tip

Remember that in regions with no charge inside, the electric field within a shell is zero due to symmetry, but the potential can still change as we move across boundaries.