Answer:
a) The charge of the capacitor is 4.25x10⁻¹¹C
b) The charge of the capacitor is 4.25x10⁻¹¹C because the battery is disconnected.
c) The potential difference across the plates is 18 V
d) The work is 7.64x10⁻¹⁰J
Explanation:
The capacitance of the capacitor is equal to:
A = 2 cm² = 0.0002 m²
d = 0.5 mm = 0.0005 m
Replacing:
a) The charge of the capacitor is equal to:
Q = C*V = 3.54 * 12 = 42.48 pC = 4.25x10⁻¹¹C
b) The charge is the same because the battery is disconnected (Q = 4.25x10⁻¹¹C)
c) If distance is increased, we have:
The potential is:
d) The work done is equal to:
