Question

Two 10.0-cm-diameter electrodes 0.50cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 14V battery.
Part A

What is the charge on each electrode while the capacitor is attached to the battery?

Enter your answers numerically separated by a comma.

Part B

What is the electric field strength inside the capacitor while the capacitor is attached to the battery?

Part D

What is the charge on each electrode after insulating handles are used to pull the electrodes away from each other until they are 1.3cm apart? The electrodes remain connected to the battery during this process.

Enter your answers numerically separated by a comma.

Part E

What is the electric field strength inside the capacitor after insulating handles are used to pull the electrodes away from each other until they are 1.3cm apart? The electrodes remain connected to the battery during this process.

Part F

What is the potential difference between the electrodes after insulating handles are used to pull the electrodes away from each other until they are 1.3cm apart? The electrodes remain connected to the battery during this process.

Answer 1

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.