A 4.0 μF parallel plate capacitor is connected to a 5 V battery. The work done by an external agent to slowly increase the separation between the plates to double of its original value is
Answers
Given info : A 4.0 μF parallel plate capacitor is connected to a 5 V battery.
To find : The work done by an external agent to slowly increase the separation between the plates to double of its original value is ..
solution : work done by an external agent = negative of change in potential energy
= initial potential energy - final potential energy
here initial potential energy, U_i = 1/2 CV²
= 1/2 × 4 × 10¯⁶ F × (5 V)²
= 2 × 10¯⁶ × 25 J
= 5 × 10¯⁵ J
if the separation between the plates is doubled, capacitance becomes half.
∵ capacitance, C = ε₀A/d , you see capacitance is inversely proportional to separation between the plates.
so new capacitance, C' = 4/2 = 2μF
now final potential energy , U_f = 1/2 × 2 × 10¯⁶ F × (5 V)²
= 1 × 10¯⁶ × 25 J
= 2.5 × 10¯⁵ J
now work done = U_i - U_f
= 5 × 10¯⁵ J - 2.5 × 10¯⁵ J
= 2.5 × 10¯⁵ J
Therefore the Workdone by an external agent is 2.5 × 10¯⁵ J.
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