Question

(II) An electric field of \({\bf{8}}{\bf{.50 \times 1}}{{\bf{0}}^{\bf{5}}}\;{\bf{V/m}}\) is desired between two parallel plates, each of area \({\bf{45}}{\bf{.0}}\;{\bf{c}}{{\bf{m}}^{\bf{2}}}\) and separated by 2.45 mm of air. What charge must be on each plate?

Short answer

The charge on each plate is\(3.39 \times {10^{ - 8}}\;{\rm{C}}\).

Step-by-step solution

Understanding electric field

The electric field between the parallel plates is equivalent to the ratio of voltage supplied to the separation distance between the plates.

The electric field between the plates is given by,

\(E = \frac{V}{d}\) …… (i)

Here, E is the electric field, V is the potential difference and d is the separation between the plates.

Given Data

The electric field between the plates is,\(E = 8.50 \times {10^5}\;{\rm{V/m}}\).

The area of each plate is,\(A = 45.0\;{\rm{c}}{{\rm{m}}^2}\).

The separation between the plates is, \(d = 2.45\;{\rm{mm}}\).

Evaluation of the charge on each plate

The charge on capacitor plate is given by,

\(\begin{aligned}Q &= CV\\Q &= \left( {\frac{{{\varepsilon _0}A}}{d}} \right)V\\Q &= {\varepsilon _0}AE\end{aligned}\)

Here, \({\varepsilon _0}\) is the permittivity of free space,Cis the capacitance andEis the electric field.

Substitute the values in the above expression.

\(\begin{aligned}Q &= \left( {8.85 \times {{10}^{ - 12}}\;{{\rm{C}}^2}{\rm{/N}} \cdot {{\rm{m}}^2}} \right)\left( {45.0\;{\rm{c}}{{\rm{m}}^2} \times \frac{{1\;{{\rm{m}}^2}}}{{{{10}^4}\;{\rm{c}}{{\rm{m}}^2}}}} \right)\left( {8.50 \times {{10}^5}\;{\rm{V/m}}} \right)\\Q &\approx 3.39 \times {10^{ - 8}}\;{\rm{C}}\end{aligned}\)

Thus, the charge on each plate is \(3.39 \times {10^{ - 8}}\;{\rm{C}}\).