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STD 12 - 2. Electric potential and capacitance — JEE STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceJEESTD 12 - 2. Electric potential and capacitance4 Marks
MCQ
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 $Q$

 if $V < CV$

  • A
    $V$
  • B
    $V + \frac{Q}{C}$
  • ✓
    $V + \frac{Q}{{2C}}$
  • D
    $V - \frac{Q}{C}$ ,
✓

Answer

Correct option: C.
$V + \frac{Q}{{2C}}$
c
After redistribution half of total charge remains on outer surface and then apply

conservation of charge on each plate.

Charge on outer plates $=\frac{Q+C V-C V}{2}=\frac{Q}{2}$

Charge of innerface of first plate,

$=Q+C V-\frac{Q}{2}=\frac{Q}{2}+C V$

Charge on innerface of second plate $=-\left(\frac{Q}{2}+C V\right)$

$V^{\prime}=\frac{Q / 2+C V}{C}=V+\frac{Q}{2 C}$

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