Question

(a) Will the electric field strength between two parallel conducting plates exceed the breakdown strength for air\(\left( {3.0 \times {{10}^6}\;V/m} \right)\)/ if the plates are separated by\(2.00\;mm\)and a potential difference of\(5.0 \times {10^3}\;V\)is applied?

(b) How close together can the plates be with this applied voltage?

Short answer

The required solutions are:

(a) The electric field between the two plates will not be stronger than the air breakdown strength\(3.0 \times {10^6}\;V/m\).

(b) The plates will be \(1.7 \times {10^{ - 3}}\;m\) close together with the applied voltage.

Step-by-step solution

Principle

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

The given data

• The breakdown strength for air is:

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

• The distance between the two plates is:

\(\begin{aligned}{c}d = (2.00\;mm)\left( {\frac{{1\;m}}{{1000\;mm}}} \right)\\ = 0.00200\;m.\end{aligned}\)

• The potential difference between the two plates is:

\(\Delta V = 5.0 \times {10^3}\;V\).

Calculation of the electric field 

(a)

Equation\((1)\)is used to calculate the electric field between the two plates:

\(E = \frac{{\Delta V}}{d}\)

The result can be obtained by plugging in the values for\(\Delta V\)and\(d\)

\(\begin{aligned}{c}E = \frac{{5.0 \times {{10}^3}}}{{0.00200\;m}}\\ = 2.5 \times {10^6}\;V/m\end{aligned}\)

Therefore, the electric field between the two plates will be \(3.0 \times {10^6}\;V/m\).

Calculation of the minimum distance between two plates 

(b)

According to Equation, the minimal distance between the two plates corresponds to the air breakdown strength \((1)\)

\({d_{\min }} = \frac{{\Delta V}}{{{E_{\max }}}}\)

Substitute the following numerical values:

\(\begin{aligned}{c}{d_{\min }} = \frac{{5.0 \times {{10}^3}\;V}}{{3.0 \times {{10}^6}\;V/m}}\\ = 1.7 \times {10^{ - 3}}\;m\end{aligned}\)

Therefore, the minimum distance between two plates is \(1.7 \times {10^{ - 3}}\;m\).