Calculating the Capacitance — Question 1

Question 1

A plate capacitance follows from charge field and voltage in sequence

Use $\epsilon_0=8.854\times10^{-12}$ F/m. Two large parallel plates of area A=.0120 m$^2$ are separated by d=.800 mm in vacuum. Neglect fringing.

1. Derive C from an assumed plate charge Q and calculate its value.

2. At voltage 150 V find the plate charge and gap field, checking both routes to E.

Question 1 - Solution

A plate capacitance follows from charge field and voltage in sequence

(a) Method and calculation

$$E=Q/(\epsilon_0A),\quad V=Ed=Qd/(\epsilon_0A).$$$$C=Q/V=\epsilon_0A/d=132.810\,\mathrm{pF}.$$

(b) Interpretation and further calculation

$$Q=CV=19.9215\,\mathrm{nC},\quad E=V/d=1.87500\times10^5\,\mathrm{V/m}.$$$$Q/(\epsilon_0A)=1.87500\times10^5\,\mathrm{V/m}.$$

Physical check

Area must be in square meters and the millimeter gap must be converted before substitution.