Q121 / 200 · NEET 2006 · Physics

A parallel plate air capacitor is charged to a potential difference of $V$ volts. After disconnecting the charging battery the distance between the plates of the capacitor is increased using an insulating handle. As a result the potential difference between the plates:

  • Correct answer
    A
    increases
  • B
    decreases
  • C
    does not change
  • D
    becomes zero

Solution

According to the question $\phi=$ Const and decrease in $C$ means increase in $V$. $$ \begin{aligned} & C=\frac{t o f}{d} \\ & \text { Capacitance }=\frac{\text { Charge }}{\text { Potential difference }} \\ & \Rightarrow \text { Potential difference } \\ & =\frac{\text { Charge }}{\text { Capacitance }} \text {. } \\ & \end{aligned} $$

Hence, the correct answer is option: 'A'.

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