Consider two concentric spherical metal shells of radii `r_(1)` and `r_(2) (r_(2) gt r_(1))`. If the outer shell has a charge q and the inner one is grounded, then the charge on the inner shell is

Options

A. `(-r_(2))/r_(1) q`
B. zero
C. `(-r_(1))/r_(2) q`
D. `-q`

Answer

To solve the problem, we need to determine the charge on the inner spherical shell when the outer shell has a charge \( q \) and the inner shell is grounded. Here’s a step-by-step solution:

### Step 1: Understand the setup
We have two concentric spherical shells:
- The inner shell (shell B) has a radius \( r_1 \) and is grounded.
- The outer shell (shell A) has a radius \( r_2 \) (where \( r_2 > r_1 \)) and has a charge \( q \).

### Step 2: Grounding condition
Since the inner shell is grounded, its electric potential \( V_B \) must be zero.

### Step 3: Calculate the potential at the inner shell
The potential \( V \) at a point due to a charged spherical shell is given by:
\[ V = \frac{kQ}{r} \]
where \( k \) is the Coulomb's constant, \( Q \) is the charge, and \( r \) is the distance from the center.

The potential at the outer shell (shell A) due to its own charge \( q \) is:
\[ V_A = \frac{kq}{r_2} \]

### Step 4: Consider the charge on the inner shell
Let \( Q' \) be the charge on the inner shell (shell B). The potential at the inner shell due to its own charge is:
\[ V_B = \frac{kQ'}{r_1} \]

### Step 5: Set up the equation for potential
Since the inner shell is grounded, we have:
\[ V_B + V_A = 0 \]
Substituting the expressions for \( V_A \) and \( V_B \):
\[ \frac{kq}{r_2} + \frac{kQ'}{r_1} = 0 \]

### Step 6: Solve for \( Q' \)
Rearranging the equation gives:
\[ \frac{kQ'}{r_1} = -\frac{kq}{r_2} \]
Dividing both sides by \( k \) (which is non-zero) results in:
\[ \frac{Q'}{r_1} = -\frac{q}{r_2} \]
Multiplying both sides by \( r_1 \) gives:
\[ Q' = -\frac{r_1}{r_2} q \]

### Conclusion
Thus, the charge on the inner shell is:
\[ Q' = -\frac{r_1}{r_2} q \]

### Final Answer
The charge on the inner shell is \( -\frac{r_1}{r_2} q \).

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