Answer:
a. 194.71 C b. 2800 V d. 74.9 C e. 1076.92 V f. The potential difference remains 14 V because the electrodes are still connected to the battery.
Explanation:
a. The charge Q, on a capacitor is Q = CV where C = capacitance and V = voltage. In a parallel plate capacitor, capacitance C = εA/d where A is the area between the plates and d is the distance between them. Since we have a diameter, d₁ = 10 cm in the question, we have circular plates and the area, A = πd²/4
So, our charge Q = εAV/d = επd₁²V/4d
Given V = 14 V, d₁ = 10 cm = 0.1 m d = 0.5 cm = 0.5 × 10⁻² m.
Q = 8.854 × 10⁻¹² × 3.142 × 0.1² × 14/(4 × 0.5 × 10⁻²) = 3.8942/0.02 = 194.71 C
b. The electric field strength E is given by E = V/d = 14/0.5 × 10⁻² = 2800 V
c. Since the distance changes to 1.3 cm apart, d₂ = 1.3 cm the charge becomes Q = επd₁²V/4d₂ = 8.854 × 10⁻¹² × 3.142 × 0.1² × 14/(4 × 1.3 × 10⁻²) = 74.9 C
e. Since the distance changes to 1.3 cm apart, d₂ = 1.3 cm, the electric field strength E is given by E = V/d₂ = 14/1.3 × 10⁻² = 1076.92 V
f. The potential difference remains 14 V because the electrodes are still connected to the battery.
