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}](https://tex.z-dn.net/?f=E%20%3D%5Cfrac%7B%20%5CDelta%20V_c%7D%7Bd%7D)
where
is the potential difference
d is the distance between the eletrode in d capacitor
= 15V
d = 0.50 cm = 0.005m
![E = \frac{15}{0.005} \\\\E = 3000V/m\\E = 3.0kV/m](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7B15%7D%7B0.005%7D%20%5C%5C%5C%5CE%20%3D%203000V%2Fm%5C%5CE%20%3D%203.0kV%2Fm)
The electric field strength inside the capacitor is 3.0kV/m
Charge on plate of the capacitor
![Q = EAe_o](https://tex.z-dn.net/?f=Q%20%3D%20EAe_o)
substitute
E = 3000V/m
![e_o =8.85 \times 10^-^1^2C^2/N.m^2](https://tex.z-dn.net/?f=e_o%20%3D8.85%20%5Ctimes%2010%5E-%5E1%5E2C%5E2%2FN.m%5E2)
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}](https://tex.z-dn.net/?f=E%20%3D%20%5Cfrac%7BV%7D%7Bd%7D)
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