CUET Preparation Today
Case : Read the passage and answer the question(s). In a parallel plate air capacitor having plate separation 0.05mm, an electric field of 4X104 V/m is established between the plates. After the removal of the battery a metal plate of thickness t = 0.02 mm is inserted between the plates of the capacitor. |
What is the potential difference across capacitor if dielectric slab with dielectric constant K=3 and same thickness were inserted in place of metal plate? |
1535 V 1500 V 1452 V 1466 V |
1466 V |
| When a dielectric slab is inserted into a capacitor, it changes the effective separation between the plates. Since the charge remains constant (battery is disconnected), the potential difference changes proportionally to the effective separation. The potential difference after inserting the dielectric is given by: \[ V' = V \cdot \frac{\text{Effective Separation with Dielectric}}{\text{Original Separation}}. \] The capacitor is divided into two regions: 1. **Vacuum**: \[ d' = d - t = 0.05 \, \text{mm} - 0.02 \, \text{mm} = 0.03 \, \text{mm} = 3 \times 10^{-5} \, \text{m}. \] 2. **Dielectric**: The effective thickness of the dielectric is reduced by a factor of \( K \): \[ \frac{t}{K} = \frac{2 \times 10^{-5}}{3} = 0.6667 \times 10^{-5} \, \text{m}. \] The total effective separation is: \[ d_{\text{eff}} = d' + \frac{t}{K} = 3 \times 10^{-5} + 0.6667 \times 10^{-5} = 3.6667 \times 10^{-5} \, \text{m}. \] The initial potential difference is: \[ V = 2000 \, \text{V}. \] The new potential difference is: \[ V' = V \cdot \frac{d_{\text{eff}}}{d}. \] Substitute the values: \[ V' = 2000 \cdot \frac{3.6667 \times 10^{-5}}{5 \times 10^{-5}}. \] Simplify: \[ \frac{3.6667 \times 10^{-5}}{5 \times 10^{-5}} = 0.7333, \] \[ V' = 2000 \cdot 0.7333 = 1466.7 \, \text{V}. \] |
