Two horizontal parallel plates each of area $500\;cm^2$ are mounted 2 mm apart in vacuum. The lower plate is earthed and the upper one is given a positive charge of $0.05\;\mu \; C$. Find the electric field intensity and the potential difference between the plates.

Given,

Area of the plates $(A) = 500 \; cm^2 = 500 * 10^{-4}\;m^2$

Distance $(d) = 2\;mm = 2 * 10^{-3}\;m$

charge $(q) = 0.05\;\mu C = 0.05 * 10^{-6}\;C$

Electric field intensity $(E) = \;?$

Electric Potential $(V) = \;?$

We have,

$E =$$\frac{\sigma}{\epsilon_0} = \frac{q}{\epsilon_0\;A} = \frac{0.05\; *\; 10^{-6}}{8.85 \;*\; 10^{-12} \;*\; 500 \;*\; 10^{-4}}$

        $= 112.9 * 10^3\; V/m$

And

$V = E \;*\; d$

    $= 112.9 * 10^3 * 2 * 10^{-3} = 226\;V$

Return to Main Menu