1

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

Four closed surfaces and corresponding charge distributions are shown below. Let the respective electric fluxes through the surfaces be ${\Phi _1},$ ${\Phi _2},$ ${\Phi _3}$ and ${\Phi _4}$. Then :
A
${\Phi _1}$ < ${\Phi _2}$ = ${\Phi _3}$ > ${\Phi _4}$
B
${\Phi _1}$ > ${\Phi _2}$ > ${\Phi _3}$ > ${\Phi _4}$
C
${\Phi _1}$ = ${\Phi _2}$ = ${\Phi _3}$ = ${\Phi _4}$
D
${\Phi _1}$ > ${\Phi _3}$ ; ${\Phi _2}$ < ${\Phi _4}$

## Explanation

Net flux through a closed surface,

$\phi$ = ${{{q_{enclose}}} \over {{\varepsilon _0}}}$

qenclosed = charge enclosed by closed surface.

For surface S1,

$\phi$1 = ${1 \over {{\varepsilon _0}}}$ (2q)

For surface S2,

$\phi$2 = ${1 \over {{\varepsilon _0}}}$ (q + q + q $-$ q) = ${1 \over {{\varepsilon _0}}}$(2q)

For Surface S3,

$\phi$3 = ${1 \over {{\varepsilon _0}}}$ (q + q) = ${1 \over {{\varepsilon _0}}}$ (2q)

For surface S4,

$\phi$4 = ${1 \over {{\varepsilon _0}}}$ (8q $-$ 2q $-$ 4q) = ${1 \over {{\varepsilon _0}}}$ (2q)

$\therefore\,\,\,$ $\phi$1 = $\phi$2 = $\phi$3 = $\phi$4
2

### 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
3

### 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$ $\left[ {{{{a^2}} \over b} - {{{b^2}} \over b} + c} \right]$

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

### 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