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=(\epsilon_0\cdot A)/(d)\cdot\Delta V \\ \sigma\text{ (charge density)}=(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=(Q\cdot d)/(\epsilon_0\cdot A) \\ \Delta V=(\sigma\cdot d)/(\epsilon_0) \\ \Delta V=(5\cdot10^(-12)\cdot4\cdot10^(-3))/(8.85\cdot10^(-12)) \\ \Delta V=(20\cdot10^(-3))/(8.85) \\ \Delta V=2.26\text{ mV} \end{gathered}`

Therefore the correct option is the third one.