A parallel plate capacitor with circular plates having radius 0.07 m and a dielectric with constant 2.2.If it is desired to store 12 $\times$ 10^-6 J in this capacitor,what must the separation distance be for a voltage of 1.5 kV?

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Quizplus · Accepted Answer

The capacitance of a parallel plate capacitor is given by C = (εA)/d, where ε is the permittivity of the dielectric, A is the area of the plates, and d is the separation distance. The energy stored in a capacitor is given by U = (1/2)CV^2, where V is the voltage across the capacitor. We can use these equations to solve for the separation distance d.C = (εA)/d = (2.2 * 8.85 x 10^-12 F/m * π * (0.07 m)^2) / d = 3.03 x 10^-11 F/dU = (1/2)CV^2 = (1/2) * 3.03 x 10^-11 F/d * (1.5 kV)^2 = 3.54 x 10^-6 J/dWe want to store 12 x 10^-6 J in the capacitor, so we can set U = 12 x 10^-6 J and solve for d:12 x 10^-6 J = 3.54 x 10^-6 J/dd = 2.8 cmTherefore, the separation distance must be 2.8 cm.

A Parallel Plate Capacitor with Circular Plates Having Radius 0 ×\times×

A Parallel Plate Capacitor with Circular Plates Having Radius 0 $\times$