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2. Electric Potential and Capacitance
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How will the voltage $(V)$ between the two plates of a parallel plate capacitor depend on the distance $(d)$ between the plates, if the charge on the capacitor remains the same?
A

B

C

D

(KVPY-2019)
Solution

$(c)$ From $C=\frac{q}{V}$
We have, $V=\frac{q}{C}=\frac{q d}{\varepsilon_{0} A}$ or $V \propto d$
Hence, graph is a straight line.
When both plates are touched $(d=0)$,
system still have some capacity $(C \neq 0)$.
So, from $C=\frac{q}{V}$ we can see that at very small seperations $V \neq \infty$ or graph does not have any values for very small $d$.
$\therefore V$ versus $d$ graph is as shown below :
Standard 12
Physics
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