Two concentric metallic spherical shells are given positive charges . Then

Options

A. the outer sphere is always at a higher potential
B. the inner sphere is always at a higher potential
C. both the spheres are at the same potential
D. no prediction can be made about their potentials unless the actual values of charges and radii are known

Explanation

To analyze the problem of two concentric metallic spherical shells with positive charges, we will follow these steps:

### Step 1: Understand the Configuration
We have two concentric metallic spherical shells. Let's denote the inner shell as Shell 1 with radius \( R_1 \) and charge \( Q_1 \), and the outer shell as Shell 2 with radius \( R_2 \) and charge \( Q_2 \). Both charges are positive.

**Hint:** Remember that metallic shells distribute charge uniformly over their surfaces.

### Step 2: Determine the Electric Field
The electric field \( E \) inside a conductor in electrostatic equilibrium is zero. Therefore, the electric field inside Shell 1 (at radius \( r < R_1 \)) is zero. For the region between the shells (i.e., \( R_1 < r < R_2 \)), the electric field due to Shell 1 can be expressed using Gauss's law.

**Hint:** Use Gauss's law to find the electric field in the region between the shells.

### Step 3: Calculate the Potential
The electric potential \( V \) at a distance \( r \) from a charged sphere is given by the formula:
\[ V = \frac{1}{4\pi\epsilon_0} \frac{Q}{r} \]
For the inner shell, the potential at its surface (at \( R_1 \)) is:
\[ V_1 = \frac{1}{4\pi\epsilon_0} \frac{Q_1}{R_1} \]
For the outer shell, the potential at its surface (at \( R_2 \)) is:
\[ V_2 = \frac{1}{4\pi\epsilon_0} \frac{Q_2}{R_2} \]

**Hint:** Remember that the potential outside a charged shell is the same as if all the charge were concentrated at the center.

### Step 4: Compare the Potentials
Since both shells are positively charged, we need to compare \( V_1 \) and \( V_2 \). The potential inside the outer shell (at \( R_2 \)) is influenced by both charges. Therefore, the total potential at \( R_2 \) is:
\[ V_{outer} = V_2 + \frac{1}{4\pi\epsilon_0} \frac{Q_1}{R_2} \]
Since \( R_1 < R_2 \), the term \( \frac{Q_1}{R_1} \) will be greater than \( \frac{Q_1}{R_2} \). Thus, the potential at the inner shell will be greater than that at the outer shell.

**Hint:** Compare the denominators when evaluating the potentials.

### Step 5: Conclusion
From the analysis, we conclude that the potential on the inner surface (Shell 1) is greater than the potential on the outer surface (Shell 2). Therefore, the inner shell is at a higher potential than the outer shell.

**Final Answer:** The potential of the inner shell is greater than that of the outer shell.