A parallel plate capacitor has a dielectric with constant 3.5.The plates are circular with radius 0.07 m,and the separation gap is 0.375 mm.What energy may be stored in this capacitor if 1250 V is the applied voltage?

Question

Quizplus · Accepted Answer

The energy stored in a capacitor is given by the equation:E = 1/2 * C * V^2where:E is the energy stored in joules (J)C is the capacitance in farads (F)V is the voltage across the capacitor in volts (V)The capacitance of a parallel plate capacitor is given by the equation:C = ε * A / dwhere:ε is the permittivity of the dielectric in farads per meter (F/m)A is the area of the plates in square meters (m^2)d is the distance between the plates in meters (m)In this problem, we are given the following information:ε = 3.5A = π * (0.07 m)^2 = 0.0154 m^2d = 0.375 mm = 0.000375 mV = 1250 VPlugging these values into the equations above, we get:C = 3.5 * 0.0154 m^2 / 0.000375 m = 1.47 x 10^-8 FE = 1/2 * 1.47 x 10^-8 F * (1250 V)^2 = 9.9 x 10^-4 JTherefore, the energy stored in the capacitor is 9.9 x 10^-4 J.

A Parallel Plate Capacitor Has a Dielectric with Constant 3 ×\times×

A Parallel Plate Capacitor Has a Dielectric with Constant 3 $\times$