Question

When the potential difference between the plates of an ideal air‐filled parallel plate capacitor is 35 v, the electric field between the plates has a strength of 750 v/m. if the plate area is 4.0 × 10-2 m2, what is the capacitance of this capacitor?

Answer 1

The difference of voltage between the two plates of the capacitor is equal to the electric field intensity E times the distance between the plates:
$\Delta V = E d$
so, from this we find d, the distance between the two plates of the capacitor:
$d= \frac{\Delta V}{E}= \frac{35.0 V}{750 V/m}=0.047 m$

And since we know also the area of the plates, we can find the capacitance:
$C=\epsilon _0 \frac{A}{d}=(8.85 \cdot 10^{-12}F/m) \frac{4.0 \cdot 10^{-2}m^2}{0.047 m}=7.53 \cdot 10^{-12} F$