A variable parallel plate capacitor consists of two metal plates with area A = 1

Question

A variable parallel plate capacitor consists of two metal plates with area A = 10-2 m2, separated from one another by a spring with unscratched length do = 1.0 mm, and stiffness constant k = 1 x 10^-4 N/m, as shown in the figure below. What is the electric field E between the plates when the capacitor is given a charge of Q? The sum of the energy stored in the electric field 1/2epsilon0E^2Ad and the energy stored in the spring 1/2k(d-d_0)^2 will reach a minimum in equilibrium. What is Q so that the two plates just barely touch one another when equilibrium is attained?

Explanation / Answer

here,

area , a = 10^-2 m^2

d0 = 0.001 m

C = a * e0/d

C = 10^-2 * 8.85 * 10^-12 /0.001

C = 8.85 * 10^-11 F

k = 1 * 10^-4 N/m

(a)

electric feild between the plated , E = V/d = (Q/C*d)

E = Q /( 8.85 * 10^-11 * 0.001)

E = 1.13 * 10^13 *Q N/C

the electric feild between the plates is 1.13 * 10^13 * Q N/C

(b)

as at the equilibrium

k * d = Q * E/2

10^-4 * 0.001 = 0.5 * 1.13 * 10^13 * Q

Q = 1.77 * 10^-20 C

the charge on each plate is 1.77 * 10^-20 C

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