A dielectric is pulled out from between the plates of a capacitor which remains

Question

A dielectric is pulled out from between the plates of a capacitor which remains connected to a battery.
Part A
What changes occur to the capacitance? (It will increase.It will decrease.It will remain the same.)
Part B
What changes occur to the charge on the plates? (It will increase.It will decrease.It will remain the same.)

Part C

What changes occur to the potential difference? (It will increase.It will decrease.It will remain the same.)
Part D
What changes occur to the energy stored in the capacitor? (It will increase.It will decrease.It will remain the same.)
Part E
What changes occur to the electric field? (It will increase.It will decrease.It will remain the same.)

Explanation / Answer

Capacitance C of a parallel palte capacitor is given by C = KeoA/d

where A = area = pi r^2, e0 = constnat = 8.85*10^-12,

d = distance between the plates, K = dieelctric constant (=1 for air)

Chareg Q = CV where V = Volatge

Energy U = 0.5QV = 0.5 CV^2 = Q^2/2C

eletric field E = V/d

PART A: if K is removed C dreases

part B: as Q = CV , charge Q reamins same

part C: V = Vo/k, V Increases

Part D: as U = Uo/K = eenrgy INcreases

part E: =EF E = Eo /K so EF increases

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