A) decreases
B) increases
C) becomes zero
D) does not change
Correct Answer: A
Solution :
Capacitance, \[C=\frac{V}{q}\] \[\left[ V = potential difference,\,\,q =charge \right]\] Also capacitance, \[C=\frac{{{\varepsilon }_{0}}A}{d}\] \[\left[ A = area of plate,d =distance between plates \right]\] As d increases capacitance decreases according to \[C=\frac{{{\varepsilon }_{0}}A}{d}\] and battery is disconnected, so charge is constant. \[\therefore \] Potential difference decreases according to\[C=\frac{V}{q}\].You need to login to perform this action.
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