A parallel plate capacitor with circular plates having radius 0.07 m is filled with a dielectric of constant 2.2. If it is desired to store 12 × 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 energy stored in a capacitor is given by:$$U = \frac{1}{2}CV^2$$where:* U 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:$$C = \frac{\epsilon_0 A}{d}$$where:* C is the capacitance in farads (F)* $\epsilon_0$ is the permittivity of free space (8.85 × 10^-12 F/m)* A is the area of the plates in square meters (m^2)* d is the distance between the plates in meters (m)We are given the following information:* U = 12 × 10^-6 J* V = 1.5 kV = 1500 V* $\epsilon_r$ = 2.2* r = 0.07 mWe can first calculate the area of the plates:$$A = \pi r^2 = \pi (0.07 m)^2 = 0.0154 m^2$$We can then calculate the capacitance:$$C = \frac{\epsilon_0 \epsilon_r A}{d} = \frac{(8.85 	imes 10^{-12} F/m)(2.2)(0.0154 m^2)}{d}$$We can then substitute the given values into the equation for energy stored:$$12 	imes 10^{-6} J = \frac{1}{2}\left(\frac{(8.85 	imes 10^{-12} F/m)(2.2)(0.0154 m^2)}{d}ight)(1500 V)^2$$Solving for d, we get:$$d = 2.8 	imes 10^{-2} m = 2.8 cm$$Therefore, the correct answer is A.

A Parallel Plate Capacitor with Circular Plates Having Radius 0