Question

Calculate the most charge that a 100pF capacitor (with a 1.0mm plate separation) can store if the capacitor breaks down with an electric field of 6.0x10 "N/C. A. 1.7nC B.0.600C C.0.30uC OD. 6.0C OE. 3.00

Answer 1

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