A dielectric slab is inserted between the plates of a capacitor. The charge on the capacitor is `Q` and the magnitude of the induced charge on each surface of the dielectric is `Q'`.

Options

A. Q' may be larger than Q.
B. Q' must be larger than Q.
C. Q' must be equal to Q.
D. Q'must be smaller tha Q.

Answer

To solve the problem, we need to analyze the situation of a parallel plate capacitor with a dielectric slab inserted between its plates. The charge on the capacitor is \( Q \), and we want to determine the relationship between the charge \( Q' \) induced on the surfaces of the dielectric slab and the charge \( Q \) on the capacitor.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- We have a parallel plate capacitor with charge \( Q \) on one plate and \( -Q \) on the other plate.
- When a dielectric slab with dielectric constant \( K \) is inserted, it affects the electric field between the plates.

2. **Electric Field Without Dielectric**:
- The electric field \( E_0 \) in the absence of the dielectric is given by:
\[ E_0 = \frac{Q}{A \epsilon_0} \]
where \( A \) is the area of the plates and \( \epsilon_0 \) is the permittivity of free space.

3. **Electric Field With Dielectric**:
- When the dielectric is inserted, the electric field \( E \) becomes:
\[ E = \frac{Q}{A \epsilon_0 K} \]
This is because the dielectric reduces the electric field by a factor of \( K \).

4. **Induced Charge on the Dielectric**:
- The dielectric slab will have induced charges on its surfaces due to the electric field. The charge induced on the positive surface of the dielectric is \( Q' \).

5. **Relation Between Charges**:
- The induced charge \( Q' \) can be related to the electric field due to polarization. The electric field due to the induced charge is:
\[ E_p = \frac{Q'}{A \epsilon_0} \]
- The net electric field \( E \) between the plates can be expressed as:
\[ E = E_0 - E_p \]

6. **Setting Up the Equation**:
- Substituting the expressions for \( E_0 \) and \( E_p \):
\[ \frac{Q}{A \epsilon_0 K} = \frac{Q}{A \epsilon_0} - \frac{Q'}{A \epsilon_0} \]
- Multiplying through by \( A \epsilon_0 \) to eliminate the denominators:
\[ \frac{Q}{K} = Q - Q' \]

7. **Solving for Induced Charge \( Q' \)**:
- Rearranging the equation gives:
\[ Q' = Q - \frac{Q}{K} \]
- Factoring out \( Q \):
\[ Q' = Q \left(1 - \frac{1}{K}\right) \]

8. **Analyzing the Result**:
- Since \( K > 1 \) (the dielectric constant is always greater than 1), the term \( \left(1 - \frac{1}{K}\right) \) is positive but less than 1. Therefore:
\[ Q' < Q \]

### Conclusion:
The induced charge \( Q' \) on each surface of the dielectric slab must be smaller than the charge \( Q \) on the capacitor. Thus, the correct answer is that \( Q' \) must be smaller than \( Q \).