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.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have two concentric spherical shells:
- Inner shell with radius \( r_1 \) (grounded).
- Outer shell with radius \( r_2 \) (charged with \( q \)).
- Grounding the inner shell means its potential is set to zero.

2. **Potential Due to the Outer Shell**:
- The potential \( V \) at a distance \( r \) from a charged spherical shell is given by the formula:
\[ V = \frac{kQ}{r} \]
where \( k \) is Coulomb's constant and \( Q \) is the charge on the shell.
- For the outer shell (with charge \( q \)), the potential at the radius \( r_1 \) (where the inner shell is located) is:
\[ V_{outer} = \frac{kq}{r_1} \]

3. **Potential of the Inner Shell**:
- Let \( Q' \) be the charge on the inner shell. The potential due to the inner shell itself at its surface (radius \( r_1 \)) is:
\[ V_{inner} = \frac{kQ'}{r_1} \]

4. **Total Potential at the Inner Shell**:
- The total potential \( V_{total} \) at the inner shell (which is grounded) must equal zero:
\[ V_{total} = V_{outer} + V_{inner} = 0 \]
- Substituting the expressions we derived:
\[ \frac{kq}{r_1} + \frac{kQ'}{r_1} = 0 \]

5. **Solving for the Charge on the Inner Shell**:
- Rearranging the equation gives:
\[ \frac{kQ'}{r_1} = -\frac{kq}{r_1} \]
- We can cancel \( k \) and \( r_1 \) (assuming \( r_1 \neq 0 \)):
\[ Q' = -q \]

6. **Conclusion**:
- The charge on the inner shell is:
\[ Q' = -q \]

### Final Answer:
The charge on the inner shell is \( -q \).