1.

A parallelplate capacitor is charged and then isolated . The effect if increasing the plate separation on charge, potential and capacitancerespectively are

Answer»

constant, increase, decrease
constant, decrease, increase
increases, decreases, decreases
constant, decreases, decreases

Solution :Since CAPACITOR is ISOLATED charge Q= Constant

Potential difference `V= EL = (sigma)/(epsi_(0)) L`
Since Lincreases, Vincreases CAPACITANCE `C= (epsi_(0)A)/(L)` as L increases C decreases
Caution: The plate separation cannot be increased indefinitely OTHERWISE `vec(E )= (sigma)/(epsi_(0)) hat(n)` can.t bemaintainced within the capacitor plates


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