Question 618: Electrostatics & Capacitance

Question

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

Accepted Answer

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
$${C_1} = {{{A \over 2}{ \in _0}} \over {2d}}$$
$$ \Rightarrow {C_1} = {{A{ \in _0}} \over {4d}}$$
$$ \Rightarrow {C_1} = {C \over 4}$$

Answer

$${C \over 4}$$

Answer

2C

Answer

$${C \over 2}$$

Answer

4C

Electrostatics & Capacitance

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

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