A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant `K= (5)/(3)` is inserted between the plates, the magnitude of the induced charge will be :

Options

A. `0.9 n C`
B. `1.2 n C`
C. `0.3 n C`
D. `2.4 n C`

Answer

To solve the problem step by step, we will follow the concepts of capacitance, charge, and the effect of inserting a dielectric material into a capacitor.

### Step-by-Step Solution:

1. **Identify Initial Parameters:**
- Capacitance of the capacitor, \( C = 90 \, \text{pF} = 90 \times 10^{-12} \, \text{F} \)
- Voltage (emf) of the battery, \( V = 20 \, \text{V} \)

2. **Calculate Initial Charge (Q):**
- The charge stored in a capacitor is given by the formula:
\[ Q = C \cdot V \]
- Substituting the values:
\[ Q = 90 \times 10^{-12} \, \text{F} \cdot 20 \, \text{V} = 1800 \times 10^{-12} \, \text{C} = 1.8 \, \text{nC} \]

3. **Determine the New Capacitance with Dielectric:**
- When a dielectric is inserted, the new capacitance \( C' \) is given by:
\[ C' = K \cdot C \]
- Where \( K = \frac{5}{3} \).
- Therefore:
\[ C' = \frac{5}{3} \cdot 90 \times 10^{-12} \, \text{F} = 150 \times 10^{-12} \, \text{F} = 150 \, \text{pF} \]

4. **Calculate New Charge (Q'):**
- The new charge with the dielectric inserted is:
\[ Q' = C' \cdot V \]
- Substituting the new capacitance:
\[ Q' = 150 \times 10^{-12} \, \text{F} \cdot 20 \, \text{V} = 3000 \times 10^{-12} \, \text{C} = 3.0 \, \text{nC} \]

5. **Calculate Induced Charge (Q_induced):**
- The induced charge on the dielectric is the difference between the new charge and the initial charge:
\[ Q_{\text{induced}} = Q' - Q \]
- Substituting the values:
\[ Q_{\text{induced}} = 3.0 \, \text{nC} - 1.8 \, \text{nC} = 1.2 \, \text{nC} \]

### Final Answer:
The magnitude of the induced charge is \( 1.2 \, \text{nC} \).

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