Question

A mica-dielectric capacitor breaks down when E volts is applied. The mica is removed and the spacing between plates doubled. If breakdown now occurs at 500 V, what is E?

Answer 1

The initial breakdown voltage of the mica-dielectric capacitor is 1000V.

Step-by-step explanation:

In this question, we are given that a mica-dielectric capacitor breaks down when E volts is applied. Then, the mica is removed and the spacing between the plates is doubled. Now, the breakdown occurs at 500V. We need to find the value of E.

Let's assume the initial breakdown voltage when the mica-dielectric is present is V1. And the breakdown voltage after removing the mica-dielectric and doubling the spacing is V2.

We can use the formula for capacitance in a parallel plate capacitor, C=ε0εrA/d, where C is the capacitance, ε0 is the permittivity of free space, εr is the relative permittivity or dielectric constant, A is the area of the plates, and d is the spacing between the plates.

Since the mica is removed and the spacing is doubled, the new spacing is 2d. The area and the dielectric constant remain the same. So, we can write:

V2 = Q/(2ε0εrA/d) = (1/2)(V1) = V1/2

Given that V2 = 500V, we can rewrite the equation as:

500 = V1/2

Solving for V1, we get:

V1 = 1000V

So, the initial breakdown voltage when the mica-dielectric is present is 1000V.