Question

When a dielectric slab is inserted between the plates of a capacitor, do the following
properties of that capacitor increase, decrease or remain the same: (a) capacitance
(b) charge (c) potential difference, and (d) potential energy (e) electric field? Justify
your answer

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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Answer 2

Answer:

answer is (b) chareges ok na