Answer:
0.6 μC
Explanation:
C = capacitance of the capacitor = 100 x 10⁻¹² F
d = separation between the plates of capacitor = 1 mm = 1 x 10⁻³ m
E = Electric field = 6 x 10⁶ N/C
Q = Amount of charge
V = Potential difference
Potential difference is given as
V = E d
Amount of charge stored is given as
Q = CV
hence
Q = C E d
inserting the values
Q = (100 x 10⁻¹²) (6 x 10⁶) (1 x 10⁻³)
Q = 6 x 10⁻⁷ C
Q = 0.6 μC
