A parallel plate capacitor consists of square plates of edge length 3.62 cm sepa

Question

A parallel plate capacitor consists of square plates of edge length 3.62 cm separated by a distance of 2.79 mm. The capacitor is charged with a 19.7-V battery, and the batten is then removed. A 2.79-mm-thick sheet of nylon (dielectric constant = 3) is slid between the plates. What is the average force (magnitude and direction) on the nylon sheet as it is inserted into the capacitor? Indicate direction as follow s: a positive sign of the force means that the direction of the force is the same direction as the motion of the dielectric material; a negative sign indicates the opposite direction.

Explanation / Answer

Capacitance C = A*eo/d
C = ((3.62 * 10^-2)^2 * 8.85 * 10^-12)/ (2.79 * 10^-3)
C = 4.156 * 10^-12 F
Q = C * V
Q = 4.156 * 10^-12 * 19.7 C

Energy E= 0.5 * Q^2/C

Now, Di electric is inserted bwteen the plates -
Cf = K*C
Vf = Q/Cf
Energy = 0.5 * Cf * Vf^2
Energy = 0.5 * K*C * Q^2/ (KC)^2
Energy = 0.5 * Q^2/KC
Ef = E/K

Reduction in Energy = E (1-1/K)

F*L = E (1-1/K)
F * (3.62 * 10^-2) = 0.5 * Q^2/C * (1-1/3)
F * (3.62 * 10^-2) = 0.5 * ((4.156 * 10^-12 * 19.7)^2/( 4.156 * 10^-12)) * (1-1/3)
F = + 1.48 * 10^-8 N

This force will pull the di-electric into the Cpacitor, i.e in the motion of the dielectric material.

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