Question

An air-filled parallel-plate capacitor has plates of area 2.40 cm^2 separated by 3.00 mm. The capacitor is connected to a(n) 23.0 V battery.

(a) Find the value of its capacitance.

pF

(b) What is the charge on the capacitor?

pC

(c) What is the magnitude of the uniform electric field between the plates?

N/C

Answer 1

a. The formula to calculate capacitance is

where k is the relative permittivity of the dielectric. For air, k=1

=8.854*10^-12 F/m (permittivity of free space)

A= 2.4 cm^2 =2.4*10^-4 m^2

d=3*10^-3m

substituting the values in the above expression, we get

C=0.708 pF

b. To find charge on the capacitor-

where Q is the charge, C is the capacitance and V is the voltage

V=23 V

C= 0.708 * 10^-14 F

Putting the values of V and C in the above equation we get,

Q=16.284 pC

c. The electric field of a capacitor is

=8.854*10^-12 F/m (permittivity of free space)

A= 2.4 cm^2 =2.4*10^-4 m^2

Q=16.284 pC= 16.284*10^-14 C

Putting the values in the above equation we get,

E=76.632 N/C

Hope this helps! Have a good day!