Question

A capacitor is constructed from two square metal plates. The gap between the plates is filled with two dielectrics of equal size as shown in the figure below. Neg fects. lect edge ef- dielectric A7 constant 4.7 left dielectric constant 74 right .4 0.12 mm Calculate the capacitance C of the device Answer in units of pF Question 4, chap 26, sect 5. part 2 of 210 points left What is the ratio -- of electric poten- right tial across the dielectric in the left-half region to that across the dielectric in the right-hal region!

A capacitor is constructed from two square metal plates. The gap between the plates is filled with two dielectrics of equal size as shown in the figure below. Neglect edge effects. 1.59 cm 0.12 mm dielectric constant right 7.4 dielectric constant left 4.7 Calculate the capacitance C of the device. Answer in units of pF.

part 2: What is the ratio Vlef t Vright of electric potential across the dielectric in the left-half region to that across the dielectric in the right-half region?

Answer 1

A = Area of each plate = 1.59 x 1.59 x 10^-4

d = width = 0.12 x 10^-3 m

Capacitance for dielectric k = 4.7 is given as

C₁ = k $\epsilon$A/d = 4.7 (8.85 x 10^-12) (1.59 x 1.59 x 10^-4) / (0.12 x 10^-3 ) = 8.76 x 10^-11 F

Capacitance for dielectric k = 7.4 is given as

C₂ = k $\epsilon$A/d = 7.4 (8.85 x 10^-12) (1.59 x 1.59 x 10^-4) / (0.12 x 10^-3 ) = 1.38 x 10^-10 F

Net Capacitance C = C₁ C₂ / (C₁ + C₂) = ( 8.76 x 10^-11 ) (1.38 x 10^-10 ) / (8.76 x 10^-11 + 1.38 x 10^-10 )

C = 5.36 x 10^-11 F = 53.6 x 10^-12 = 53.6 pF

Q_total = CV

V_left = Q_total /C₁

V_right = Q_total /C₂

Taking the ratio

V_left / V_right = C₂ / C₁

V_left / V_right = (1.38 x 10^-10 ) / (8.76 x 10^-11 )

V_left / V_right = 1.58