Question

Two 2.50 cm × 2.50 cm plates that form a parallel-plate capacitor are charged to ± 0.708 nC .

Part A

What is the electric field strength inside the capacitor if the spacing between the plates is 1.50 mm ?

Part C

What is the electric field strength inside the capacitor if the spacing between the plates is 3.00 mm ?

Answer 1

Part A)

A = area of each plate of the capacitor = 2.50 x 2.50 = 6.25 x 10^-4 m²

Q = magnitude of charge stored on each plate = 0.708 x 10^-9 C

V = potential difference between the plates = ?

d = distance between the plates = 1.50 mm = 1.50 x 10^-3 m

Capacitance is given as

C = A/d

Potential difference between the plates is given as

Q = CV

Q = AV/d

V = Qd/( A)

V/d = Q/( A)

E = Q/( A)

inserting the values

E = (0.708 x 10^-9)/((8.85 x 10^-12) (6.25 x 10^-4))

E = 1.28 x 10⁵ N/C

Part C)

A = area of each plate of the capacitor = 2.50 x 2.50 = 6.25 x 10^-4 m²

Q = magnitude of charge stored on each plate = 0.708 x 10^-9 C

V = potential difference between the plates = ?

d = distance between the plates = 3 mm = 3 x 10^-3 m

Capacitance is given as

C = A/d

Potential difference between the plates is given as

Q = CV

Q = AV/d

V = Qd/( A)

V/d = Q/( A)

E = Q/( A)

inserting the values

E = (0.708 x 10^-9)/((8.85 x 10^-12) (6.25 x 10^-4))

E = 1.28 x 10⁵ N/C