A parallel-plate capacitor in air has a plate separation of 1.49 cm and a plate

Question

A parallel-plate capacitor in air has a plate separation of 1.49 cm and a plate area of 25.0 cm2. The plates are charged to a potential difference of 275 V and disconnected from the source. The capacitor is then immersed in distilled water. Assume the liquid is an insulator. (a) Determine the charge on the plates before and after immersion. before pC after pC (b) Determine the capacitance and potential difference after immersion. Cf = F ?Vf = V (c) Determine the change in energy of the capacitor. nJ

Explanation / Answer

a) Q = CV = e0 A V/d = 8.85E-12*25.0E-4*275/1.49E-2=4.08E-10 C = 408 pC won't change with immersion since charge has nowhere to go b) Not sure which value to use for the dielectric constant, Im going to use 80 but check what your book uses C = k e0A/d =8.85E-12*25.0E-4*80/1.49E-2=1.188E-10 F V =Q/C = 408E-12/1.188E-10=3.43 V c) dE = 1/2 Cf Vf^2 - 1/2 Ci Vi^2 =0.5*1.188E-10*3.43^2- 0.5* (8.85E-12*25.0E-4*275^2/1.49E-2)=-55.4 nJ

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