A parallel-plate capacitor in air has a plate separation of 1.50 cm and area of
ID: 2063139 • Letter: A
Question
A parallel-plate capacitor in air has a plate separation of 1.50 cm and area of 25.0 cm^2. The plates are charged to a potential difference of 250 V and disconnected from the source. The capacitor is then immersed in distilled water. Determine:(a) the charge on the plates before and after immersion,
(b) the capacitance and potential difference after immersion, and
(c) the change in the amount of energy stored in the capacitor.
Assume the liquid is an insulator.
Explanation / Answer
since the capacitor is disconnected the charge remains the same. a ) the charge before and after immersion is q = c*V where c= e*A/d = 1.475*10^-12 F = 1.475 pF => q = 250*c = 0.368 nC b) the capacitance is C' = k*c = 76.7*c = 113.1325 pF ( i have taken the dielectric constant of water to be 76.7 ) => the potential difference is V' = v/k = 3.259 V c) the energy change is given by E = 4.530*10^-8 J