NEETPhysicsCapacitorMCQ+4 / −1
The distance between the two plates of a parallel plate capacitor is doubled and the area of each plate is halved. If C is its initial capacitance, its final capacitance is equal to
- A
- B2C
- C
- D4C
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Correct answer: A
For a capacitor of area A and distance between the plates as d, the capacitance (C) is
$$C = {{A{ \in _0}} \over d}$$
On doubling the distance and reducing area to half
$${C_1} = {{{A \over 2}{ \in _0}} \over {2d}}$$
$$ \Rightarrow {C_1} = {{A{ \in _0}} \over {4d}}$$
$$ \Rightarrow {C_1} = {C \over 4}$$
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