Q19.2-21PE - College Physics (Urone)

Question

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 \(3.0 \times {10^6}{\rm{ }}V/m\).

Step-by-Step Solution

Answer

Value of the maximum potential difference between the two parallel plates is\(\Delta {V_{\max }} = 15.0 \times {10^3}\;V\).

Step 1: Relation between the electric field and potential difference and calculation of the electric field.

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

Step 2: The data given

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

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

Step 3: Calculation for the maximum potential difference between the plates.

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 \(\Delta {V_{\max }} = 15.0 \times {10^3}\;V\).