Question

A capacitor of capacitance $C$ is charged to a potential difference $V$ from a cell and then disconnected from it. A charge $ + Q$ is now given to its positive plate. The potential difference across the capacitor is now:
A. $V$
B. $v + \dfrac{Q}{C}$
C. $V + \dfrac{Q}{{2C}}$
D. $V - \dfrac{Q}{C},$ if $Q < CV$

Answer 1

Hint:
Here we need to first find the potential difference and then the electric fields for both the positive and negative charge. At last we have to find the total potential difference.

Complete step by step answer:
Given,
Capacitance of the capacitor $ = C$
Potential difference of a cell $ = V$
Charge given on the positive plate $ = + Q$
Let the initial charge on the positive plate be ${Q_ \circ }$
So, total charge on the positive plate $ = Q + {Q_ \circ }$
Let the charge on the negative plate be $ - {Q_ \circ }$
Let the new potential difference after giving new charge be $V'$
We know that,
Charge on a parallel plate capacitor is capacitance times potential difference
${Q_ \circ } = CV$
The new potential difference will be
 $
  V' = Ed \\
   = ({E_1} + {E_2})d \\
 $

Where, ${E_1}$
 is the electric field due to charge $Q + {Q_ \circ }$ and ${E_2}$ is the electric field due to charge $ - {Q_ \circ }$
$d$ is the gap between the two plates.
Thus, the electric fields are given by-
${E_1} = \dfrac{{Q + {Q_ \circ }}}
{{2{\varepsilon _ \circ }A}}$
${E_2} = \dfrac{{{Q_ \circ }}}
{{2{\varepsilon _ \circ }A}}$
Thus,
$
  V' = \left( {\dfrac{{Q + {Q_ \circ }}}{{2{\varepsilon _ \circ }A}} + \dfrac{{{Q_ \circ }}}{{2{\varepsilon _ \circ }A}}} \right)d \\
   = \dfrac{{Qd}}{{2{\varepsilon _ \circ }A}} + \dfrac{{{Q_ \circ }d}}{{{\varepsilon _ \circ }A}} \\
   = \dfrac{Q}{{2C}} + V \\
$
Hence, the potential difference across the capacitor is now: $\dfrac{Q}{{2C}} + V$

So, the correct answer is “Option C”.

Additional Information:
In a parallel plate capacitor if an uncharged capacitor $C$ is attached through the terminals of a potential $V$ battery, then when the capacitor plates are charged, a transient current flows. As soon as the charge $Q$ on the positive plate crosses the value $Q = CV$, the current ceases flowing.
The electric field between the capacitor plates cancels the influence of the electric field produced by the battery when the charge exceeds the $Q = CV$ value, resulting in no further movement.
Therefore, when attached with a battery, it conducts for a while and then operates as an open circuit.
Charge is supplied to the capacitor plates while a capacitor is attached to a battery. In this phase, the battery needs to operate to provide the capacitor with electrical energy, and this energy is deposited between the plates in the form of an electrostatic field.
In the electrical field between its plates, a charged capacitor stores electricity. The electrical field builds up as the capacitor is being charged. As a charged capacitor is removed from a battery, the energy in the gap between its plates remains in the area.

Note:
Here we have to see that although the charge is negative we have not given the negative sign while calculating the potential difference. Also there would be two electric fields for the two plates of the capacitor.