Question

The plates of a parallel-plate capacitor are 3.50 mm apart, and each carries a charge of magnitude 75.0 nC. The plates are in vacuum. The electric field between the plates has a magnitude of 5.00×10^6 V/m.a. What is the potential difference between the plates?b. What is the area of each plate?c. What is the capacitance?

Answer 1

Vab =17.5kV

A = 16.9 cm2

C = 4.27pF

Step-by-step explanation:

a) Find the voltage difference:

Vab = Ed

E Electric field

d distance between plates

Vab potential difference

d = 3.5mm

= 3.5 * 10^(-3) m

Q = 75.0nC

= 75 * 10^(-9)

E = 5.00 * 10^6 V/m

Vab = (5.00 * 10^6) * (3.5 * 10^(-3))

= 17.5 * 10^3 V

=17.5kV

b. What is the area of the plate?

The relation between the electric field and area is given as:

E = Q/(ϵ0 * A)

A = Q/(ϵ0 *E)

Where ϵ0 is the electric constant and equals 8.854 × 10^ (-12) C2/N•m2

A = 75 * 10^ (-9) / (8.854 × 10^ (-12) (5.00 * 10^6)

= 1.69 X 10^ (-3) m2

= 16.9 cm2

c. Find the capacitance

The equation relating capacitance, area of plate and plate distance is given by:

C = ϵ0 A/d

plug in the values of d, ϵ0 and A above to get the capacitance:

C = (8.854 × 10^ (-12) * 1.69 X 10^ (-3) / 3.5 * 10^ (-3)

= 4.27 * 10^ (-12) F

= 4.27pF