Question

A parallel plate capacitor consists of two circular plates of radius 10 cm that are spaced 2 mm apart with air between the plates . The electric field between the plates is 2.0x105 V/m . What is the capacitance

Answer 1

The capacitance of the parallel-plate capacitor depends on the area $A=\pi R^2$ of plates, the distance between plates $d$ and the permittivity of medium between plates $\varepsilon$ (see https://en.wikipedia.org/wiki/Capacitor#Parallel-plate_capacitor or http://hyperphysics.phy-astr.gsu.edu/hbase/electric/pplate.html), so we can write

$C = \dfrac{\varepsilon A}{d} = \dfrac{\varepsilon \pi R^2}{d}.$ The permittivity of air is approximately $\varepsilon_0 = 8.854\cdot10^{-12}\,\mathrm{F/m}$ , so the capacitance will be

$C = \dfrac{8.854\cdot10^{-12}\,\mathrm{F/m}\cdot \pi \cdot (0.1\,\mathrm{m})^2}{0.002\,\mathrm{m}} =1.39\cdot10^{-10}\,\mathrm{F} .$

The charge that is stored in the capacitor, is

$Q = CEd = 1.39\cdot10^{-10}\mathrm{F}\cdot2.0\cdot10^5\,\mathrm{V/m}\cdot0.002\,\mathrm{m} = 5.56\cdot10^{-8}\,\mathrm{C}.$

The voltage drop will be

$\Delta V = Ed = 2.0\cdot10^5\,\mathrm{V/m}\cdot0.002\,\mathrm{m} = 4\cdot10^2\,\mathrm{V}.$