Question

The electric field between the plates of a paper-separated (K=3.75) capacitor is 8.26×104 V/m . The plates are 2.20 mm apart, and the charge on each plate is 0.685 μC .

A. Determine the capacitance of this capacitor.

Express your answer using three significant figures and include the appropriate units.

B. Determine the area of each plate.

Express your answer using three significant figures and include the appropriate units.

Answer 1

here,

dielectric constant , K = 3.75

electric feild , E = 8.26 * 10^4 V/m

the separation between the plates , d = 2.2 mm

d = 2.2 * 10^-3 m

charge , Q = 0.685 uC = 0.685 * 10^-6 C

a)

the capacitance , C = Q /V

C = Q /( E * d)

C = 0.685 * 10^-6 /(8.26 * 10^4 * 2.2 * 10^-3) C

C = 3.77 * 10^-9 F

the capacitance is 3.77 * 10^-9 F

b)

let the area of each plate be A

C = K * A * e0 /d

3.77 * 10^-9 = 3.75 * A * 8.85 * 10^-12 /(2.2 * 10^-3)

solving for A

A = 0.25 m^2

the area of each plate is 0.25 m^2