To design the parallel plate capacitor with the given specifications, we need to determine the minimum area of the plates. Let's break down the given information and use it step by step.
The capacitor has a dielectric constant
κ=5 and a dielectric strength of
106Vm−1. The voltage rating is 2 kV and the maximum electric field allowed is
10% of its dielectric strength.
First, calculate the maximum electric field:
Emax=0.1×106Vm−1=105Vm−1Given the voltage rating, the separation between the plates (d) can be found using the relation:
V=Emax×d Rearranging for
d :
d=‌=‌=0.02mNext, we use the formula for the capacitance of a parallel plate capacitor:
C=κE0‌ Where:
‌C=60‌pF=60×10−12F‌κ=5‌E0=8.85×10−12Fm−1‌d=0.02m‌A=‌ Area of the plates ‌Rearranging for
A :
A=‌Substituting the values:
Simplifying further, we get:
A≈2.71×10−2m2Therefore, the minimum area of the plates should be:
Option B:
2.7×10−2m2