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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorForce & Energy analysisEasy2 minQB

PhysicsEasysingle choice

A capacitor with capacitance $5 μ F$ is charged to $5 μ C$ . If the plates are pulled apart to reduce the capacitance to $2 μ F$ , how much work is done?

Options:

Answer:

Solution:

$U_{i n i t i a l} = \frac{Q^{2}}{2 C_{1}}$
$U_{f i n a l} = \frac{Q^{2}}{2 C_{2}}$
Work done $= U_{f i n a l} - U_{i n i t i a l}$
$= \frac{Q^{2}}{2 C_{2}} - \frac{Q^{2}}{2 C_{1}}$
$= \frac{{(5 \times 10^{- 6})}^{2}}{2 \times 2 \times 10^{- 6}} - \frac{{(5 \times 10^{- 6})}^{2}}{2 \times 5 \times 10^{- 6}}$
$= (\frac{25}{4} - \frac{5}{2}) \times 10^{- 6}$
$= \frac{15}{4} \times 10^{- 6} = 3.75 \times 10^{- 6} J$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Force & Energy analysis

⏱ 2mℹ️ Source: QB

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