Question

a capacitor is filled with an insulator and a certain potential difference is applied to its plates. the energy stored in the capacitor is u. now the capacitor is disconnected from the source and the insulator is pulled out of the capacitor. the work done against the forces of electric field in pulling out the insulator is 4u. then dielectric constant of the insulator is

Please answer the question with explanation.
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Answer 1

$\huge\mathtt\pink{A}\mathtt\red{N}\mathtt\blue{S}\mathtt\green{W}\mathtt\purple{E}\mathtt\green{R} \: 5$

FANTASTIC question

Explanation:-

we know that value of potential energy can be described as F.dr ( force x DISPLACEMENT)

AND we know that force applied on the slab at a distance x is =

$\large \: f = \frac{ {q}^{2}d(k - 1) }{2 \epsilon_ol( {l + (k - 1)x})^{2} }$

So the total energy required to pull the slab is F.dr which is :-

$\large \delta u = f. \delta x$

$\large u = \int_{0}^{l}\frac{ {q}^{2}d(k - 1) }{2 \epsilon_ol( {l + (k - 1)x})^{2} }. \delta x$

$u = \int_{0}^{l} \frac{ {q}^{2} d(k - 1)}{2 \epsilon_ol} \times ( \frac{ -1 }{l + (k - 1)x} )( \frac{1}{k - 1} )$

$u = \int_{0}^{l} \frac{ {q}^{2} d}{2 \epsilon_ol} \times ( \frac{ -1 }{l + (k - 1)x} )( \frac{1}{1} )$

$u = \frac{ {q}^{2} d}{2 \epsilon_ol} \times ( \frac{ -1 }{l + (k - 1)l } + \frac{1}{l} )$

$u = \frac{ {q}^{2} d}{2 \epsilon_o {l}^{2} } \times ( 1 - \frac{1}{k} )$

Now this is the energy required to pull out the slab.

And according to question energy of capacitor is 4 times of this energy

Energy of capacitor:-

$\large\frac{ {q}^{2} }{2ck} = \frac{ {q}^{2}d }{2 \epsilon_oa}$

here a = area of plate = lxl

equate both energies

$4\times \frac{ {q}^{2} d}{2 \epsilon_oak} = \frac{ {q}^{2} d}{2 \epsilon_o {l}^{2} } \times ( 1 - \frac{1}{k} )$

$\frac{4}{ k} = ( 1 - \frac{1}{k} )$

$\frac{5}{k} = 1$

$k = 5$

NOTE :- in last fourth step value of a is l²

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