Question

The battery remains connected to a parallel plate capacitor and a dielectric slab is inserted between the plates. What will be the effect on itsi. capacityii. chargeiii. potential differenceiv. electric fieldv. energy stored?

Answer 1

1- Capacity Capacity increases C =KC↓0
2 - Potential V remains same V = V ↓0
3 - Charge Q remains same Q = CV
4 - Electric field E Decreases E = E↓0/k
5 - energy stored in the condenser increase U = 1/2 CV sq. = KU↓0

Answer 2

Hey !!

When a battery remains connected

• Potential difference V remains constant.
• Capacity C increases.
• Electric field will remains same.
• Energy stored 1/2 CV² increase as C increase.

DETAILED EXPLANATION

When a battery remains connected to a parallel plate capacitor and if a dielectric slab is inserted between the plates of capacitor, then

(1) There will be no change in the potential difference as the capacitor remained connected with the battery.

(2) Capacity or capacitance will increase since with the introduction of dielectric slab, capacitance of capacitor will result C = Kε₀A/d where k > 1 resulting an increase in C.

(3) Electric field will remains same as there will be no change in potential difference and distances between the plates.

(4) Energy stored will be increased since from the expression E = 1/2CV², potential difference V remains same while C increases which finally increases the energy of capacitor.

GOOD LUCK !!