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PhysicsCapacitorBasic capacitor analysis with dielectricMedium2 minQB

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Figure shows two identical capacitors A & B with identical dielectric inserted between them and are connected to a battery. Now the slab of capacitor B is pulled out with battery connected :

Options:

Answer:

A, C, D

Solution:

$C_{i} = \frac{CK}{2} q_{i} = \frac{CK}{2} E$

$C_{f} = \frac{CC K}{C + C K} = \frac{C K}{1 + K}$

$q_{f} = \frac{CKE}{1 + K} q_{f} < q_{i}$ meano change will flow from capacitor to battery

$EF_{i} = \frac{E}{2 d}$

$E F_{f} = \frac{\frac{C K E}{1 + K}}{K C d} = \frac{E}{( 1 + K ) d}$

Stream:JEESubject:PhysicsTopic:CapacitorSubtopic:Basic capacitor analysis with dielectric

⏱ 2mℹ️ Source: QB

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