Question

A parallel plate capacitor , each with plate area a, separation between the plate is d, is charged to a potential difference v. The battery used to charge remains connected, and a dielectric slab of dielectric constant k is now placed between the plates, what change if any will takes place in ? (a)charge on the plates? (b) electric field intensity between the plates (c) capacitance of capacitor. Justify your answer in each case

Answer 1

Answer:

(i) The capacitance increases as the dielectric constant K>1.

(ii) Potential difference V=

C

Q

. As C increases and Q remains the same since the battery is disconnected, the p.d. between the plates decreases.

(iii) Electric field E=

d

V

where V is the p.d. and d the separation between the plates. As V decreases and d remains the same, electric field also decreases.

(iv) Energy stored in a capacitor U=

2

1

C

Q

2

. As Q is constant and C increases, U decreases.

Explanation:

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