Question

A parallel-plate capacitor has plates with an area of 1.0×10⁻² m2 and a separation of 0.82 mm . The space between the plates is filled with a dielectric whose dielectric constant is 2.1.

A. What is the potential difference between the plates when the charge on the capacitor plates is 4.2 μC ?

B. Will your answer to part A increase, decrease, or stay the same if the dielectric constant is increased?

Answer 1

here,

the area of each plate , A = 1 * 10^-2 m^2

speration , d = 0.82 mm = 8.2 * 10^-4 m

dielectric constant , K = 2.1

the capacitance , C = K * area * e0 /d

C = 2.1 * 1 * 10^-2 * 8.85 * 10-12 /(8.2 * 10^-4)

C = 2.27 * 10^-10 F

a)

the potential difference , V = Q /C

V = 4.2 * 10^-6 /( 2.27 * 10^-10)

V = 1.85 * 10^4 V

b)

capacitance , C = K * area * e0 /d

when the dielectric constant increases

the capacitance increases

and

V = Q/C

as capacitance increases , the potential difference decreases

so, the answer to part A decreased