Q17PE.
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?
Step-by-Step Solution
VerifiedThe 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.
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 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:
\(\Delta V = 5.0 \times {10^3}\;V\).
(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 \(3.0 \times {10^6}\;V/m\).
(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 \(1.7 \times {10^{ - 3}}\;m\).