need help with capacitance 12) A parallel-plate capacitor has plates of area A =

Question

need help with capacitance

12) A parallel-plate capacitor has plates of area A = 250 cm2 and separation d = 2.00 mm. The capacitor is charged to a potential difference Vo 150 V. Then the battery is disconnected (the charge Q on the plates then won't change), and a dielectric sheet (K-3.50) of the same area A but thickness l =1.00 mm is centered between the plates. Determine: (a)the initial capacitance of the air-filled capacitor; is inserted; (c)the charge induced on each face of the dielectric after it is inserted; (d)the electric field in the space between each plate and the dielectric; (e)the electric field in the dielectric; (f)the potential difference between the plates after the dielectric is added; and (g)the capacitance after the dielectric is in place le)the capacitance after the

Explanation / Answer

(A) C = e0 A / d

C = (8.854 x 10^-12) (250 x 10^-4) / (2 x 10^-3)

C = 1.107 x 10^-10 F

(B) Q = C V = 1.107 x 10^-10 x 150 = 1.66 x 10^-8 C

(C) Q' = Q / k = (1.66 x 10^-8) / 3.50 = 4.74 x 10^-9 C

(D) E = Q / A e0 = (1.66 x 10^-8)/(250 x 10^-4 x 8.854 x 10^-12)

E = 75000 N/C

(E) E' = Q' / Ae0 = 21428.57 N/C

(F) V' = E (d -t) + E' t

= (75000 x (0.002 - 0.001)) + (21428.57 x 0.001)

= 96.43 Volt

(g) Q = C' V'

C' = (1.66 x 10^-8)/ 96.43 = 1.72 x 10^-10 F

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