A parallel plate capacitor has a capacitance of 10 microfarads. If the distance between the plates is doubled and the area of the plates is halved, what is the new capacitance?

A. 2.5 F — The capacitance of a parallel plate capacitor is proportional to the area and inversely proportional to the distance. Changing these parameters results in a new capacitance calculated by the ratio of the changes.

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$A parallel plate capacitor has a capacitance C = 10 μ F . If the distance between the plates d is doubled and the area of the plates A is halved, what is the new capacitance C^{'} ?$

A. $2.5 μ F$
B. $5 μ F$
C. $10 μ F$
D. $20 μ F$

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Correct answer: A — $2.5 μ F$

Solution

The capacitance is C = \frac{ϵ _{0} A}{d} . The new capacitance is C^{'} = \frac{ϵ _{0} ( A /2 )}{2 d} = \frac{1}{4} C . Thus, C^{'} = 10/4 = 2.5 μ F .

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A parallel plate capacitor has a capacitance C=10 μF. If the distance between the plates d is doubled and the area of the plates A is halved, what is the new capacitance C′?

Solution

Attempt this question & track your score

More Electrostatics practice questions

$A parallel plate capacitor has a capacitance C = 10 μ F . If the distance between the plates d is doubled and the area of the plates A is halved, what is the new capacitance C^{'} ?$