Question

A charge of 1μC is given to one plate of a parallel-plate capacitor of capacitance 0.1μF and a charge of 2μC is given to the other plate. Find the potential difference developed between the plates.

Answer 1

Hint: Here we have a parallel plate capacitor and, on each plate, there are two different sets of charge. When the current goes through the parallel plate capacitor, after some time equilibrium is reached and thus, we have to find the equilibrium charge which can be found by adding the two charges and dividing by two.

Complete step by step answer:
\[{{q}_{1}}=1\mu C\]
\[{{q}_{2}}=2\mu C\]
So net charge on the capacitor is \[q=\dfrac{{{q}_{1}}-{{q}_{2}}}{2}=-0.5\times {{10}^{-6}}C\]
Now the capacitance of the plate is 0.1μF, that is \[0.1\times {{10}^{-6}}F\]
We know for a parallel plate capacitor, q=CV
\[\begin{align}
  & V=\dfrac{q}{C} \\
 & \Rightarrow \dfrac{-0.5\times {{10}^{-6}}}{0.1\times {{10}^{-6}}} \\
 & \therefore -5V \\
\end{align}\]
But potential can never be (–) negative. So, V = 5 V

Additional Information:
when the capacitor is connected to a battery then the charging process gets started. The charge moves from one plate of the capacitor to the other and so an electric field is produced in the space between the two plates. We know capacitance of a parallel plate capacitor is given by \[C=\dfrac{{{\varepsilon }_{0}}A}{d}\]

Note:
While putting the values we have to keep in mind that the units are taken in standard SI. Also, while finding the net charge we see it is coming out to be negative and that can be possible. A plate if negatively charged it means the direction of the electric field is towards the plate. Also, any extra charge is possible resides on the surface and not inside the plate.