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 ?

A parallel plate capacitor is charged with a battery and then separate

1.	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 ?
Answer» remains constant, decreases, decreases increases, decreases, decreases remains constant, decreases, increases remains constant, increases, decreases Solution :In absence of BATTERY, charge on the plates of capacitor remains the same. Hence, ACCORDING to V = Ed, V Increases andfrom ` C = (epsilon_(0)A)/(d)` C decreases as d increases.

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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 ?

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