Question

Find capacitance of parallel plate capacitor with square plates. Length of the side is 10 cm and distance between plates is 2 mm. Find charge on the plates and electrical energy stored in the capacitor if it is connected to the battery providing 250 V. Find capacitance, energy and charge if capacitor filled with dielectric with K=3 (still connected to the same battery).

Answer #1

Here, initially

capacitance of parallel plate capacitor = epsilon * Area/d

capacitance of parallel plate capacitor = 8.854 *10^-12 * 0.10^2/0.002

capacitance of parallel plate capacitor = 4.43 *10^-11 F

-------------

change on plates = 250 * 4.43 *10^-11

change on plates = 1.11 *10^-8 C

-------------

energy stored = 0.5 * 4.43 *10^-11 * 250^2

energy stored = 1.38 *10^-6 J

----------------------------------

Now, when a new dielectric is entered

capacitance = 3 * 4.43 *10^-11 = 1.329 *10^-10 F

charge on plate = 3 * 1.11 *10^-8 C = 3.33 *10^-8 C

energy stored = 3 * 1.38 *10^-6 = 4.14 *10^-6 J

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