Start Free Mock LoginAlready registered? Login here

Physics - Electrostatics Question with Solution | TestHub

PhysicsElectrostaticsMiscellaneousEasy2 minPYQ_2020

PhysicsEasysingle choice

In the given circuit switch $K$ is open. The charge on the capacitor is $C$ in steady-state is $q_{1}$ . Now the key is closed and steady-state charge on $C$ is $q_{2}$ . The ratio of charges $q_{1} / q_{2}$ is -

Options:

Answer:

Solution:

Initially when switch is open

Finally when the switch is closed

$I = \frac{E}{3 R}; ∴ V_{A} - V_{B} = I \times 2 R = E / 3 R \times 2 R = 2 E / 3$

$∴ C h a r g e o n c a p a c i t o r = C V; q_{2} = \frac{C 2 E}{3}$

$∴ \frac{q_{1}}{q_{2}} = \frac{C E}{2 / 3 C E} = 3 : 2$

Stream:NTA_ABHYAS_NEETSubject:PhysicsTopic:ElectrostaticsSubtopic:Miscellaneous

⏱ 2mℹ️ Source: PYQ_2020

Doubts & Discussion

Loading discussions...