Question

The capacitance of a parallel plate capacitor is 12 uF. If the
distance between the plates is doubled and area is halved,
then new capacitance will be
(a) 8 F
(b) 6uF
(c) 4uF
(d) 3uF​

Answer 1

Answer:

3ųF

Explanation:

If you have any doubts, please ask

Answer 2

Answer:

(d)The new capacitance is 3uF

Explanation:

Given that,

The capacitance of Parallel plate capacitor=12uF

The relation between Capacitance,Area of plate and distance between plates is given as

$C = ε _{0} \frac{A}{d}$

Let the initial distance=d

Therefore,New distance=2d

Let the initial area=A

Therefore,New Area=0.5A

$C ∝\frac{A}{d} \\ \frac{C _{1} }{C _{2} } = \frac{A _{1} }{A _{2} } \times \frac{d _{2} }{d _{1}}$

Substituting Known values we get,

$\frac{12}{C _{2} } = \frac{A}{0.5A} \times \frac{2d}{d} \\ C _{2} = 3uF$

Therefore,the new capacitance is 3uF(d)

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