Question

A parallel plate capacitor consists of two plates of $2m\times 1m$. The space between the plates is of 1mm and found with a dielectric of relative permittivity of 7. A potential difference of 300 volts is applied across the plates. Find the potential gradient.

$\begin{align}
  & \text{A}\text{. 6}\times \text{1}{{\text{0}}^{5}}N/C \\
 & \text{B}\text{. 3}\times \text{1}{{\text{0}}^{5}}N/C \\
 & \text{C}\text{. 18}\times \text{1}{{\text{0}}^{5}}N/C \\
 & \text{D}\text{. 12}\times \text{1}{{\text{0}}^{5}}N/C \\
\end{align}$

Answer 1

Hint: In the question, we have to find the values that are given. Then at first, we will have to find the capacitance of the capacitor. Then with that capacitance, we have to find the charge of the capacitor and after finding the charge we have to find the charge density, and then only we can find the potential gradient of the charge.

Formulas used:
$\text{capacitance=}\dfrac{K{{\xi }_{\circ }}A}{d}$
Q=CV
$\sigma =\dfrac{\text{charge}}{\text{area}}$
$E=\dfrac{\sigma }{2{{\xi }_{\circ }}}$

Complete step by step answer:
We know that the area of the plate is $2m\times 1m$which is $2{{m}^{2}}$
And the distance between the plates is (d) = 1mm or ${{10}^{-3}}m$
And we also know that the potential difference is (V) = 300V.
So according to the problem we have to find the capacitance first,
$\text{capacitance=}\dfrac{K{{\xi }_{\circ }}A}{d}$
\[C=\dfrac{7\times 8.85\times {{10}^{-12}}\times 2}{{{10}^{-3}}}\] ,
On solving this we get,
\[C=123.9\times {{10}^{-9}}\].
We know that,
Q=CV
\[Q=123.9\times {{10}^{-9}}\times 300\]
\[Q=371.70\times {{10}^{-7}}\].
We know that charge density for a capacitor is,
$\sigma =\dfrac{\text{charge}}{\text{area}}$
$\sigma =\dfrac{371.70\times {{10}^{-7}}}{\text{2}}$
$\sigma =185.85\times {{10}^{-7}}$,
Therefore the potential gradient of the electric field must be,
$E=\dfrac{\sigma }{2{{\xi }_{\circ }}}$ ,
$E=\dfrac{185.85\times {{10}^{-7}}}{2\times 8.85\times {{10}^{-12}}}$ ,
Which on solving gives,
$E=10.5\times {{10}^{5}}N/C$.
Therefore option D is the correct option.

Note:
We are considering option D as the correct option as it is the nearest to the answer and all the other options are too far from the result. Remember to convert all the units into the SI system. The question is solved in many parts. Many formulas were used and the chances of mistakes are more so try to look after the formula one or more times and try to solve it with patience. In the equation $\sigma =\dfrac{\text{charge}}{\text{area}}$, $\sigma $ is the charge density of the capacitor.