The figure shows a parallel-plate capacitor of plate area A and plate separation

Question

The figure shows a parallel-plate capacitor of plate area A and plate separation d. A potential difference V0 is applied between the plates. While the battery remains connected, a dielectric slab of thickness b and dielectric constant is placed between the plates as shown. Assume A = 129 cm2, d = 1.48 cm, V0 = 87.6 V, b = 0.616 cm, and = 4.72. Calculate (a) the capacitance, (b) the charge on the capacitor plates, (c) the electric field in the gap, and (d) the electric field in the slab, after the slab is in place.

Explanation / Answer

A) Capacitance C= Aeo /(d - b+b/K)

= 129e-4*8.85e-12/(0.0148-0.00616+0.00616/4.72)

= 1.148*10^-11 F

B) Q = CV

= 1.148e-11*87.6

= 1.006*10^-9 C

C) Electric field E = sigma/eo

= 1.006e-9/(8.85e-12*129e-4)

= 8811 N/C

D) electric field E = sigma/(Ke0)

= 1.006e-9/(4.72*8.85e-12*129e-4)

= 1867 N/C

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