Question 5 of 8: Minimum Plate Area for a Dielectric-Filled Capacitor at the Breakdown Limit

Question 5: Minimum Plate Area for a Dielectric-Filled Capacitor at the Breakdown Limit (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Find. The minimum plate area $A$ that keeps the field at or below $E_{\max}$ while storing $U$.

Approach. The maximum field fixes the maximum voltage $V_{\max}=E_{\max}d$; express the stored energy in terms of that voltage and the (unknown) capacitance $U=\tfrac12CV_{\max}^2$, substitute $C=\varepsilon_r\varepsilon_0A/d$, and solve for $A$ — the SMALLEST area that can hold $1$J without exceeding the field limit.

1. Express stored energy at the field limit in terms of area. At the maximum allowable field, $V_{\max}=E_{\max}d$ and $C=\varepsilon_r\varepsilon_0A/d$, so $$U=\frac12CV_{\max}^2=\frac12\left(\frac{\varepsilon_r\varepsilon_0A}{d}\right)(E_{\max}d)^2=\frac12\varepsilon_r\varepsilon_0AdE_{\max}^2$$
2. Solve for the minimum area. A smaller $A$ would force a higher field (for the same $U$) than $E_{\max}$ allows, so this is the minimum: $$A=\frac{2U}{\varepsilon_r\varepsilon_0dE_{\max}^2}=\frac{2(1)}{(2.5)(8.85\times10^{-12})(5\times10^{-4})(10^7)^2}$$ $$A=\boxed{1.808\ \text{m}^2}$$