Question

Two 10-cm-diameter electrodes 0.50 cm apart form a parallel-plate capacitor. The electrodes are attached by metal wires to the terminals of a 15 V battery. After a long time, the capacitor is disconnected from the battery but is not discharged. What are the charge on each electrode, the electric field strength inside the capacitor, and the potential difference between the electrodes

Answer 1

Answer:

a)  the charge on each plate of the capacitor is  2.1 × 10⁻¹⁰C

b) the electric field strength inside the capacitor is 1.5kV/m

c)Potential difference = 15V

Explanation:

The electric field E between the plates of capacitor when they are seperated by a distance d is equal to the ratio of the potential difference V between the plates of the capacitor to the seperation between the capacitors d

E = V/d

The electric field strength inside the capacitor is equal to the ratio of the potential difference V between the plates of the capacitor

$E =\frac{ \Delta V_c}{d}$

where

$\Delta V_c$ is the potential difference

d is the distance between the eletrode in d capacitor

$\Delta V_c$ = 15V

d = 0.50 cm = 0.005m

$E = \frac{15}{0.005} \\\\E = 3000V/m\\E = 3.0kV/m$

The electric field strength inside the capacitor is 3.0kV/m

Charge on plate of the capacitor

$Q = EAe_o$

substitute

E = 3000V/m

$e_o =8.85 \times 10^-^1^2C^2/N.m^2$

A = πr²

r = 0.005m

A = π(0.005)²

A = 7.85 × 10⁻³m

so,

Q = (3000) ×( 7.85 × 10⁻³) × (8.85 × 10⁻¹²)

Q = 2.1 × 10⁻¹⁰C

Therefore, the charge on each plate of the capacitor is  2.1 × 10⁻¹⁰C

b)The Electric field strength inside the capacitor

$E = \frac{V}{d}$

V = 15V

d = 1cm = 0.01m

E = 15/00.1

E = 1500V/m

E = 1.5kV/m

Therefore, the electric field strength inside the capacitor is 1.5kV/m

The potential difference

After the electrodes of the capacitor are pulled away to a distance of seperation of 1.0cm, the potential difference across the capacitor remain unchanged

ΔV = 15V

Potential difference = 15V