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
[tex]E =\frac{ \Delta V_c}{d}[/tex]
where
[tex]\Delta V_c[/tex] is the potential difference
d is the distance between the eletrode in d capacitor
[tex]\Delta V_c[/tex] = 15V
d = 0.50 cm = 0.005m
[tex]E = \frac{15}{0.005} \\\\E = 3000V/m\\E = 3.0kV/m[/tex]
The electric field strength inside the capacitor is 3.0kV/m
Charge on plate of the capacitor
[tex]Q = EAe_o[/tex]
substitute
E = 3000V/m
[tex]e_o =8.85 \times 10^-^1^2C^2/N.m^2[/tex]
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
[tex]E = \frac{V}{d}[/tex]
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
