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### JEE Main 2017 (Online) 9th April Morning Slot

A combination of parallel plate capacitors is maintained at a certain potential difference.

When a 3 mm thick slab is introduced between all the plates, in order to maintain the same potential difference, the distance between the plates is increased by 2.4 mm. Find the dielectric constant of the slab.
A
3
B
4
C
5
D
6
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### JEE Main 2018 (Offline)

Three concentric metal shells A, B and C of respective radii a, b and c (a < b < c) have surface charge densities $+ \sigma$, $- \sigma$ and $+ \sigma$ respectively. The potential of shell B is :
A
${\sigma \over { \in {}_0}}\left[ {{{{b^2} - {c^2}} \over c} + a} \right]$
B
${\sigma \over { \in {}_0}}\left[ {{{{a^2} - {b^2}} \over a} + c} \right]$
C
${\sigma \over { \in {}_0}}\left[ {{{{a^2} - {b^2}} \over b} + c} \right]$
D
${\sigma \over { \in {}_0}}\left[ {{{{b^2} - {c^2}} \over b} + a} \right]$

## Explanation

Let charge of shell A, B and C are QA, QB and QC respectively.

Potential of B shell will be due to charge QA, QB and QC.

Here charge QA is inside of the shell B and QB is on the surface of the shell B in both cases you have to take the radius of the shell B, while calculating potential of shell B.

Charge QC is outside of the shell B so take radius of shell C for calculating potential of shell B.

$\therefore$ VB = V$_{{Q_a}}$ + V$_{{Q_b}}$ + V$_{{Q_c}}$

= ${1 \over {4\pi {\varepsilon _0}}}$ $\left[ {{{4\pi {a^2}\left( { + \sigma } \right)} \over b} + {{4\pi {b^2}\left( { - \sigma } \right)} \over b} + {{4\pi {c^2}\left( { + \sigma } \right)} \over c}} \right]$

= ${1 \over {4\pi {\varepsilon _0}}}$ $\times$ 4$\pi $$\sigma \left[ {{{{a^2}} \over b} - {{{b^2}} \over b} + c} \right] = {\sigma \over {{\varepsilon _0}}}\left[ {{{{a^2} - {b^2}} \over b} + c} \right] 3 MCQ (Single Correct Answer) ### JEE Main 2018 (Offline) A parallel plate capacitor of capacitance 90 pF is connected to a battery of emf 20 V. If a dielectric material of dielectric constant K = 5/3 is inserted between the plates, the magnitude of the induced charge will be : A 0.9 n C B 1.2 n C C 0.3 n C D 2.4 n C ## Explanation Charge on Capacitor initially, Qi = CV After inserting dielectric of dielectric constant = K, new capacitance, Qf = (KC) \vee \therefore\,\,\, Induced charges on dielectric = Qf - Qi = KCV - CV = (K - ) CV = \left( {{5 \over 3} - 1} \right) \times 90 \times 10-12 \times 20 = 1.2 \times 10-9 C = 1.2 nC 4 MCQ (Single Correct Answer) ### JEE Main 2018 (Online) 15th April Morning Slot A body of mass M and charge q is connected to spring of spring constant k. It is oscillating along x-direction about its equilibrium position, taken to be at x=0, with an amplitude A. An electric field E is applied along the x-direction. Which of the following statements is correct ? A The new equilibrium position is at a distance {{qE} \over {2k}} from x=0. B The total energy of the system is {1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}{{{q^2}{E^2}} \over k}. C The total energy of the system is {1 \over 2}m{\omega ^2}{A^2} - {1 \over 2}{{{q^2}{E^2}} \over k}. D The new equilibrium position is at a distance {{2qE} \over k} from x=0. ## Explanation On the body of charge q a electric fied E is applied, because of this equilibrium position of body will shift to a point where resulttant force is zero. \therefore\,\,\,\, kxeq = qE \Rightarrow$$\,\,\,$ xeq = ${{qE} \over K}$

$\therefore\,\,\,\,$ Total energy of the system

= ${1 \over 2}m{\omega ^2}{A^2} + {1 \over 2}K\,x_{eq}^2$

= ${1 \over 2}$ m${\omega ^2}$A2 + ${1 \over 2}.{{{q^2}{E^2}} \over K}$