A parallel plate capacitor is charged with a battery and then separated from it. Now if the distance between its two plates is increased, what will be the changes in electric charge, potential difference and capacitance respectively ?

Options

A. remains constant, decreases, decreases
B. increases, decreases, decreases
C. remains constant, decreases, increases
D. remains constant, increases, decreases

Answer

A, C, D

Explanation

In absence of battery, charge on the plates of capacitor remains the same.
Hence, according to V = Ed, V Increases and from ` C = (epsilon_(0)A)/(d)` C decreases as d increases.