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A parallel plate condenser with a dielectric of dielectric constant K between the plates has a capacity C and is charged to a potential V volt. The dielectric slab is slowly removed from between the plates and then reinserted. The net work done by the system in this process isA. zeroB. `(1)/(2) (K - 1) CV^(2)`C. `(CV^(2) (K - 1))/(K)`D. `(K - 1) CV^(2)` |
Answer» Correct Answer - A Initial energy `U_(i) = (Q^(2))/(2 K C)` When the slab is removed, `U = (Q^(2))/(2C)`, When the slab is removed, `U = (Q^(2))/(2C)`, When dielectric slab is introduced, energy becomes `U_(f) = (Q^(2))/(2 KC) = U_(i)`. `:.` Net work done by the system `= U_(f) - U_(i)` = Zero |
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