A parallel plate capacitor has a plate area of 0.500 m^2 and a plate separation

Question

A parallel plate capacitor has a plate area of 0.500 m^2 and a plate separation of 0.100 mm. The charge on each plate has a magnitude of 5.00E-6 C. Find; a) the capacitance of the capacitor. b) the electric field (magnitude and direction) between the plates. c) the potential difference across the plates. The capacitor is now connected in parallel (there is no battery, the first capacitor is all that charges the second capacitor) to an identical capacitor. Find; d) the equivalent capacitance. e) the total charge stored in the two capacitors. A slab of porcelain, with dielectric constant K_e = 6.50 is inserted between the plates of each capacitor and the original potential difference is maintained. Find; f) the total energy stored in the two capacitors.

Explanation / Answer

(A) C = e0 A / d

= (8.854 x 10^-12) (0.500) / (0.10 x 10^-3)

= 4.427 x 10^-8 F

(B) E = sigma / e0 = (Q / A) / e0

E = (5 x 10^-6) / (8.854 x 10^-12 x 0.50)

= 1.129 x 10^6 N/C

negative charged plate to positively charge plate

(C) V= Q / C = 112.94 Volt

(D) Ceq = C1 + C2 = 2 C = 8.85 x 10^-8 F

(E) from charge conservcation, Q_total = 5 x 10^-6 C

(f) V' = (5 x 10^-6 / 8.85 x 10^-8) = 56.5 Volt

C' = 6.50C

Ceq = 2 C' = 5.755 x 10^-7 F

U_total = Ceq V'^2 /2 = 9.19 x 10^-4 J

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