A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are

Question

A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are (A) constant, decreases, decreases (B) increases, decreases, decreases (C) constant, decreases, increases (D) constant, increases, decreases

Tardigrade Admin · Accepted Answer

After separation charge is constant. Capacity, C=(ε0A/d) Capacitance decreases with increase in distance and V=(Q/C) Potential increases with decrease in capacitance (C).

Q. A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are

constant, decreases, decreases

increases, decreases, decreases

constant, decreases, increases

constant, increases, decreases

After separation charge is constant.
Capacity, $C = \frac{ε _{0} A}{d}$
Capacitance decreases with increase in distance and $V = \frac{Q}{C}$
Potential increases with decrease in capacitance (C).

Q. A parallel plate capacitor is charged and then isolated. The effect of increasing the plate separation on charge, potential and capacitance respectively are

constant, decreases, decreases

increases, decreases, decreases

constant, decreases, increases

constant, increases, decreases

After separation charge is constant. Capacity, C=dε0​A​ Capacitance decreases with increase in distance and V=CQ​ Potential increases with decrease in capacitance (C).

After separation charge is constant.
Capacity, $C = \frac{ε _{0} A}{d}$
Capacitance decreases with increase in distance and $V = \frac{Q}{C}$
Potential increases with decrease in capacitance (C).