A parallel-plate capacitor has capacitance C = 12.5 pF when the volume between t

Question

A parallel-plate capacitor has capacitance C = 12.5 pF when the volume between the plates is filled with air. The plates are circular, with radius 3.00 cm. The capacitor is connected to a battery and a charge of magnitude 25.0 pC goes onto each plate. With the capacitor still connected to the battery, a slab of dielectric is inserted between the plates, completely filling the space between the plates. After the dielectric has been inserted the charge on each plate has magnitude 45.0 pC. What is the dielectric constant K of the dielectric? What is the potential difference between the plates before the dielectric has been inserted? What is the potential difference between the plates after the dielectric has been inserted? What is the electric field at a point between the plates before the dielectric has been inserted? What is the electric field at a point between the plates after the dielectric has been inserted?

Explanation / Answer

(A) Q = C V

initially: 25 pC = (12.5 pF) (V)

V = 2 volt

As capacitor is conncted to the battery hence V will constant .

finally: Q' = C' V

C' = (45 / 2) = 22.5 pC

and C' = k C

k = 22.5 / 12.5 = 1.8

(B) V = 2 Volt

(C) V = 2 volt ( constant)

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

= (25 x 10^-12) / (pi x 0.03^2 x 8.854 x 10^-12)

= 998.64 N/C

(E) E = V/d { V and d both are constant hence E will be constant }

hence E = 998.64 N/C Or V/m

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