Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorForce & Energy analysisEasy2 minQB

PhysicsEasysingle choice

If the charge on a capacitor is increased by $2 C$ , the energy stored in it increases by $44 %$ . The original charge on the capacitor is (in $C$ )

Answer:

Solution:

Energy stored in a capacitor is given by $E = \frac{Q^{2}}{2 C}$ .

Let initial charge be $Q_{0}$ .

Then, $1.44 (\frac{Q_{0}^{2}}{2 C}) = \frac{{(Q_{0} + 2)}^{2}}{2 C}$

$\Rightarrow 1.2 Q_{0} = Q_{0} + 2$

$Q_{0} = \frac{2}{0.2} = 10 C$

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

⏱ 2mℹ️ Source: QB

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