The figure below shows a parallel plate capacitor of plate area A = 140 cm^2 and

Question

The figure below shows a parallel plate capacitor of plate area A = 140 cm^2 and plate separation d = 1.30 cm. A potential difference of V_0 = 82.0 V is applied between the plates. Suppose that the battery remains connected while the dielectric slab of thickness b = 0.780 cm and dielectric constant? = 3.60 is being introduced. Calculate the following values. the capacitance pF the charge on the capacitor plates nC the electric field in the gap N/C the electric field in the slab, after the slab is in place N/C

Explanation / Answer

capacitance C = k*e0*A/d

k = dielectric constant

e0 = permitivitty of the free space

A = area of cross section

d = seperation beween the plates

(a0

C = 3.6*8.85*10^-12*140*10^-4/(1.3*10^-2)

C = 34.3 pF <<<<=====answer

(b)

Q = C*v = 34.4*10^-12*82 = 2.821 nC

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(c)

E0 = V/d = 82/(1.3*10^-2) = 6308 n/c

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e = K*e0 = 3.6*6308 = 22709 N/C

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