Consider a parallel plate capacitor with plate dimensions of 10 cm x 10 cm and a

Question

Consider a parallel plate capacitor with plate dimensions of 10 cm x 10 cm and a gap spacing of 2.00 cm. The capacitor is connected to a battery supplying a voltage across the capacitor of 2.00 kV. After the capacitor is fully charged, the battery is disconnected. A dielectric material is then inserted into the vacuum space to completely fill the capacitor. This result in a decrease of the capacitor voltage to 1.00 kV.

a. Find the dielectric constant of the dielectric material.

b. Find the permitivity of the dielectric material.

c. Find the magnitude of the induced charge on each face of the dielectric material facing the capcitor plates.

Explanation / Answer

Here ,

A) new potential across Vp = 2 kV - 1 kV

Vp = 1 kV

dielectric constant = V/Vp

dielectric constant = 2 kV/1 kV

dielectric constant = 2

the dielectric constant of the material is 2

B)

permitivity of the dielectric material = k * epsilon0

permitivity of the dielectric material = 2 * 8.854 *10^-12

permitivity of the dielectric material = 1.77 *10^-11 F/m

the permitivity of the dielectric material is 1.77 *10^-11 F/m

C)

initial charge , q = 8.854 *10^-12 * 0.10^2 * 2000/0.02

q = 8.854 *10^-9 C

as the charge on the capacitor is constant

the charge induced on the dielectric is 8.854 *10^-9 C

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