Question

What is the potential difference across a parallel-plate capacitor whose plates are separated by a distance of 4.0 mm where each plate has a charge density of magnitude 5.0 pC/m^2?1.02 millivolts1.43 millivolts2.26 millivolts3.34 millivolts4.43 millivolts

Answer 1

In order to calculate the potential difference between the plates, we can use the following formula:

$\begin{gathered} Q=\frac{\epsilon_0\cdot A}{d}\cdot\Delta V \\ \sigma\text{ (charge density)}=\frac{Q}{A} \end{gathered}$

Using d = 4 * 10^-3 m, sigma = σ * 10^-12 and ε0 = 8.85 * 10^-12, we have:

$\begin{gathered} \Delta V=\frac{Q\cdot d}{\epsilon_0\cdot A} \\ \Delta V=\frac{\sigma\cdot d}{\epsilon_0} \\ \Delta V=\frac{5\cdot10^{-12}\cdot4\cdot10^{-3}}{8.85\cdot10^{-12}} \\ \Delta V=\frac{20\cdot10^{-3}}{8.85} \\ \Delta V=2.26\text{ mV} \end{gathered}$

Therefore the correct option is the third one.