Physics - Electrostatics Question with Solution | TestHub

As the potential of the leftmost plate is zero, we can use Gauss's law to derive the given options.

Three large parallel conducting plates are arranged as shown. The **leftmost plate is grounded**, the middle plate is neutral, and the rightmost plate has charge $-2Q$.

For a conductor in electrostatic equilibrium, the electric field inside the material is zero.

### (R) Charge on left surface of middle plate

The rightmost plate has charge $-2Q$. Since the system is grounded, no charge is required on its outer surface. Hence, its inner surface carries:

$$-2Q$$

To make the electric field inside the middle conductor zero, the left surface of the middle plate must carry:

$$\boxed{+2Q}$$

Therefore,

$$\boxed{R\to (4)}$$

### (S) Charge on right surface of middle plate

The middle plate is neutral, so the total charge on it must be zero:

$$q_{left}+q_{right}=0$$

$$2Q+q_{right}=0$$

$$q_{right}=\boxed{-2Q}$$

Therefore,

$$\boxed{S\to (3)}$$

### (P) Charge on outer surface of rightmost plate

Since the left plate is grounded, the electric field outside the system on the right side is zero. Hence, no charge appears on the outer surface of the rightmost plate:

$$\boxed{P\to (1)}$$

### (Q) Charge on outer surface of leftmost plate

The charge induced on the right surface of the grounded plate is $-2Q$. To maintain zero electric field inside the grounded conductor, the outer surface acquires:

$$\boxed{+Q}$$

Thus,

$$\boxed{Q\to (2)}$$

### Final Answer

$$\boxed{P-1,Q-2,R-4,S-3}$$