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Physics - Capacitor Question with Solution | TestHub

PhysicsCapacitorBasic capacitor analysis with dielectricEasy2 minQB

PhysicsEasysingle choice

The plates of a parallel plate capacitor are charged to a potential of $200 V$ . Now, a dielectric slab of thickness $4 mm$ is inserted between its plates and to maintain the same potential difference between the plates of the capacitor, the distance between the plates is increased by $3.2 mm$ . The dielectric constant of the slab is

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Solution:

$\frac{ε_{0} A}{d} = \frac{ε_{0} A}{d^{'} - t + \frac{t}{K}}$
$\Rightarrow d = d^{'} - t + \frac{t}{K}$
$\Rightarrow d^{'} - d = t (1 - \frac{1}{K})$

Here, $d^{'} - d = 3 .2 mm, t = 4 mm$
$∴ 3 .2 = 4 (1 - \frac{1}{K})$
$\Rightarrow \frac{3 .2}{4} = 1 - \frac{1}{K}$
$\Rightarrow 1 - \frac{1}{K} = \frac{4}{5}$
$∴ K = 5$

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

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