A parallel plate capacitor of capacitance 6.76 mu F has the space between the pl

Question

A parallel plate capacitor of capacitance 6.76 mu F has the space between the plates filled with a slab of glass with kappa = 3.23. The capacitor is charged by attaching it to a 7.5-V battery. After the capacitor is disconnected from the battery, the dielectric slab is removed. (a) Find the capacitance after the glass is removed. mu F (b) Find the potential difference after the glass is removed. V (c) Find the charge on the plates after the glass is removed. mu C (d) Find the energy stored in the capacitor after the glass is removed. mu J

Explanation / Answer

Given,

C = 6.76 uF ; k = 3.23 ; V = 7.5 V

a)after the glass is removed, C becomes:

C' = C/k = 6.76 uF/3.23 = 2.093 uF

Hence, C' = 2.093 uF (utp 3 decimal places)

b) The voltage increases.

V = Q/C

V' = Q/C'

Q remains same since the charge is conserved.

V/V' = C'/C

V' = C/C' V = 6.76/2.093 x 7.5 = 24.224 Volts

Hence, V' = 24.224 Volts

c) Charge is conserved.

Q = CV = 6.76 x 7.5 = 50.700 uF

Hence, Q = 50.700 uF

d)U = 1/2 C' V'^2

U = 1/2 x 2.093 x 10^-6 x 24.224^2 = 6.141 x 10^-4 J

Hence, U = 6.141 x 10^-4

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