If the electric field inside a capacitor exceeds the dielectric strength of the

Question

If the electric field inside a capacitor exceeds the dielectric strength of the dielectric between its plates, the dielectric will break down, discharging and ruining the capacitor. Thus, the dielectric strength is the maximum magnitude that the electric field can have without break-down occurring. The dielectric strength of air is 3.0 x 106 V/m, and that of neoprene rubber is 1.2 x 107 V/m. A certain air-gap, parallel-plate capacitor can store no more than 0.048 J of electrical energy before breaking down. How much energy can this capacitor store without breaking down after the gap between its plates is filled with neoprene rubber? Take the dielectric constant of air to be 1.0, and of neoprene rubber to be 6.7.

Explanation / Answer

Energy stored in a capacitor is 0.5 * C * V^2.
Charge in a capacitor is C * V.
Capacitance is eo * er * A / d

We can see that the difference in capacitance will be 6.7/1.0 times the air capacitor.
The voltage difference will be 1.2E7 / 3E6.

Using the energy formula, we see that energy is proportional to capacitance and V^2.
So multiply the air cap's energy by the voltage difference squared times the capacitance difference.

You should end up with E = 0.048J * (1.2E7/3E6)^2 * (6.7/1.0)= 5.14

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