Question

Two large metal plates of area $\mathbf{1}\mathbf{.}\mathbf{0}{\mathit{m}}^{\mathbf{2}}$face each other, $\mathbf{5}\mathbf{.}\mathbf{0}\mathit{c}\mathit{m}$apart, with equal charge magnitudes but opposite signs. The field magnitude$\mathbf{\left|}\mathbf{q}\mathbf{\right|}$ between them (neglect fringing) is 5N/C . Find $\mathbf{\left|}\mathbf{q}\mathbf{\right|}$.

Short answer

The value of the magnitude of the charge $\left|q\right|$is $4.9\times {10}^{-10}C$.

Step-by-step solution

The given data

a) Area of the metal plate, $A=1m^2$

b) Separation of the metal plates, d = 5cm

c) Magnitude of the electric field neglecting friction,$\left|E\right|=55\mathrm{N}/\mathrm{C}$

Understanding the concept of the electric flux

Using the concept of the electric flux passing through a closed surface, we can get the value of the magnitude of the charge by substituting the given values.

Formula:

The electric flux within a closed surface enclosing the volume,$\mathrm{E}·\mathrm{A}=\frac{\mathrm{q}}{{\mathrm{\epsilon }}_0}$ (1)

Calculation of the magnitude of the charge

Substituting the given values in equation (1), the magnitude of the charge can be calculated as follows:

$$\begin{gathered} \left|\mathrm{q}\right|=\left|\mathrm{E}\right|{\mathrm{A\epsilon }}_0 \\ =(8.85\times {10}^{-12}{\mathrm{C}}^2/\mathrm{N}·{\mathrm{m}}^2)(55\mathrm{N}/\mathrm{C})(1.0{\mathrm{m}}^2) \\ =4.9\times {10}^{-10}C \end{gathered}$$

Hence, the value of the charge is $=4.9\times {10}^{-10}C$.