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= (\Delta V)/(E)= (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 (A)/(d)=(8.85 \cdot 10^(-12)F/m) (4.0 \cdot 10^(-2)m^2)/(0.047 m)=7.53 \cdot 10^(-12) F`