A parallel plate capacitor with circular plates of radius 10.0 cm are separated by a distance of 5.0 mm.The potential across the capacitor is increased at a constant rate.If the magnitude of the magnetic field between the plates at a distance r = 1.0 cm from the center is 3.6 * 10^-15 T,what is the rate of increase in the potential across the plates of the capacitor?

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Quizplus · Accepted Answer

The rate of change of magnetic field is given by:$$\frac{dB}{dt} = \mu_0 \epsilon_0 \frac{dV}{dt}$$where:* dB/dt is the rate of change of magnetic field* $\mu_0$ is the vacuum permeability* $\epsilon_0$ is the vacuum permittivity* dV/dt is the rate of change of potentialWe are given:* dB/dt = 3.6 * 10^-15 T/s* $\mu_0$ = 4$\pi$ * 10^-7 T m/A* $\epsilon_0$ = 8.85 * 10^-12 F/m* r = 1.0 cm = 0.01 mSolving for dV/dt, we get:$$\frac{dV}{dt} = \frac{dB/dt}{\mu_0 \epsilon_0} = \frac{3.6 * 10^-15 T/s}{4\pi * 10^-7 T m/A * 8.85 * 10^-12 F/m} = 320 V/s$$Therefore, the rate of increase in the potential across the plates of the capacitor is 320 V/s.

A Parallel Plate Capacitor with Circular Plates of Radius 10