Find the maximum potential difference between two parallel conducting plates separated by \(0.500{\rm{ }}cm\) of air, given the maximum sustainable electric field strength in air to be \(3.0 \times {10^6}{\rm{ }}V/m\).

The potential difference between two points separated by a distance\(d\)in a homogeneous electric field of magnitude\(E\)is\(\Delta V = Ed{\rm{ }}......(1)\).

Now calculation for the distance between the plates is:

\(\begin{array}{c}d = (0.500\;cm)\left( {\frac{{1\;m}}{{100\;cm}}} \right)\\d = 5.00 \times {10^{ - 3}}\;m\end{array}\)

And here maximum sustainable electric field is

\({E_{\max }} = 3.0 \times {10^6}\;V/m\)

The maximum potential difference between the two plates corresponds to the maximum sustainable electric field strength in air and the distance between the plates as found from equation:

\(\begin{array}{c}\Delta {V_{\max }} = {E_{\max }}d\\\Delta {V_{\max }} = \left( {3.0 \times {{10}^6}\;V/m} \right)\left( {5.00 \times {{10}^{ - 3}}\;m} \right)\\\Delta {V_{\max }} = 15.0 \times {10^3}\;V\end{array}\)

Therefore, the maximum potential difference value is obtained as \(\Delta {V_{\max }} = 15.0 \times {10^3}\;V\).