A parallel-plate capacitor in air has a plate separation of 1.76 cm and a plate

Question

A parallel-plate capacitor in air has a plate separation of 1.76 cm and a plate area of 25.0 cm^2. The plates are charged to a potential difference of 230 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. Determine the charge on the plates before and after immersion, before before pC after pC Determine the capacitance and potential difference after immersion. C_f = F Delta V_f = V Determine the change in energy of the capacitor, nJ

Explanation / Answer

part a )

charge remain same after and before immersion

Ci = eo*A/d = 8.85 x 10^-12 * 25 x 10^-4 /1.76 x 10^-2

Ci = 1.257 x 10^-12 F

Qi = Ci*V = 289.11 pC

part b )

water dielectric constant = 80

Cf = k*Ci = 1.0056 x 10^-10 F

Vf = Q/Cf = 2.875 V

part c )

Ui = 1/2 * Ci * Vi^2

Uf = 1/2 * Cf*Vf^2

Cf = k*Ci

Vf = Vi/k

Uf = 1/2 * Ci * Vi^2/k

dU = 1/2 * Ci * Vi^2(1/k -1)

dU = 3.28 x 10^-8 J = 32.8 nJ

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