The plates of a parallel plate capacitor in vacuum are 5.0 mm apart. A potential difference of 10.0 kV is applied across them. The magnitude of the surface charge density on each plate is
Answer & explanation
Correct answer: option 1
The correct answer is Option (1) → \( 1.77\times10^{-5}\,\mathrm{C\,m^{-2}} \)
In a parallel plate capacitor in vacuum, the field between the plates is uniform, \( E=V/d \). This field is linked to the surface charge density of the plates by \( E=\sigma/\varepsilon_0 \), with \( \varepsilon_0=8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}} \).
In SI units the data are:
\( V=10.0\,\mathrm{kV}=1.0\times10^{4}\,\mathrm{V} \)
\( d=5.0\,\mathrm{mm}=5.0\times10^{-3}\,\mathrm{m} \)
The field between the plates is:
\( \begin{aligned} E &= \frac{V}{d} \\ &= \frac{1.0\times10^{4}\,\mathrm{V}}{5.0\times10^{-3}\,\mathrm{m}} \\ &= 2.0\times10^{6}\,\mathrm{V\,m^{-1}} \end{aligned} \)
The surface charge density is then:
\( \begin{aligned} \sigma &= \varepsilon_0E \\ &= 8.85\times10^{-12}\,\mathrm{C^{2}\,N^{-1}\,m^{-2}}\times2.0\times10^{6}\,\mathrm{V\,m^{-1}} \\ &= 1.77\times10^{-5}\,\mathrm{C\,m^{-2}} \end{aligned} \)