A parallel plate capacitor is connected to a battery with an EMF of V and is in
ID: 1457786 • Letter: A
Question
A parallel plate capacitor is connected to a battery with an EMF of V and is in equilibrium. A dielectric slab of dielectric constant K is slowly inserted between the plates with the battery still connected and the system is allowed to come to equilibrium again. (a) For each of the following quantities, state whether inserting the dielectric slab causes it to increase, stay the same or decrease. (i) The charge on the capacitor plates. (ii) The capacitance. (iii) The potential difference between the plates. (iv) The magnitude of the electric field. (v) The stored energy. (vi) The magnitude of the electric displacement vector.Explanation / Answer
i) The charge increases after inserting the dielectric slab since battery remains connected.
ii) The capacitance increases by the factor K, the dielectric constant.
iii) Since the battery remains connected, the potential difference remains same even after inserting dielectric.
iv) Since the voltage remains same, electric field also remains same for same separation between the plates.
v) The stored energy increases as E = (1/2)CV2
vi) The magnitude of electric displacement vector increases.