How will the voltage (V) between the two plat | vedclass

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(C) For a parallel plate capacitor,the capacitance is given by $C = \frac{\varepsilon_0 A}{d}$.Given that the charge $q$ on the capacitor remains cons

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(C) For a parallel plate capacitor,the capacitance is given by $C = \frac{\varepsilon_0 A}{d}$.Given that the charge $q$ on the capacitor remains constant,the potential difference $V$ between the plates is given by $V = \frac{q}{C}$.Substituting the expression for $C$,we get $V = \frac{q d}{\varepsilon_0 A}$.Since $q$,$\varepsilon_0$,and $A$ are constants,we have $V \propto d$.This implies that the graph of $V$ versus $d$ is a straight line passing through the origin. However,in a real physical capacitor,when the plates are in contact $(d=0)$,the capacitance is not infinite,and the potential difference is not zero due to the finite size of the plates and edge effects. Therefore,the graph starts from a small positive value of $d$ and is a straight line,which does not pass through the origin.

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?

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