1

### JEE Main 2017 (Online) 8th April Morning Slot

There is a uniform electrostatic field in a region. The potential at various points on a small sphere centred at $P,$ in the region, is found to vary between the limits 589.0 V to 589.8 V. What is the potential at a point on the sphere whose radius vector makes an angle of 60o with the direction of the field ?
A
589.5 V
B
589.2 V
C
589.4 V
D
589.6 V

## Explanation

$\Delta$V = E. d

$\Rightarrow $$\,\,\, 589.8 - 589.0 = (E d)max \Rightarrow$$\,\,\,$ (E d)max = 0.8

$\therefore\,\,\,$ $\Delta$V = E d cos$\theta$

= 0.8 $\times$ cos60o

= 0.4

$\therefore\,\,\,$ Maximum potential on the sphere = 589.4 V
2

### JEE Main 2017 (Online) 8th April Morning Slot

The energy stored in the electric field produced by a metal sphere is 4.5 J. If the sphere contains 4 $\mu$C charge, its radius will be :
[ Take : ${1 \over {4\,\pi { \in _0}}} =$ 9 $\times$ 109 N $-$ m2/C2 ]
A
20 mm
B
32 mm
C
28 mm
D
16 mm

## Explanation

Energy of sphere = ${{{Q^2}} \over {2C}}$

$\therefore\,\,\,$ ${{16 \times {{10}^{ - 12}}} \over {2C}}$ = 4.5

$\Rightarrow $$\,\,\, C = {{16 \times {{10}^{ - 12}}} \over 9} We know capacity of spherical conductor, C = 4\pi$$\varepsilon$0R

$\therefore\,\,\,$ 4$\pi $$\varepsilon 0R = {{16 \times {{10}^{ - 12}}} \over 9} \Rightarrow$$\,\,\,$ R = ${1 \over {4\pi {\varepsilon _0}}} \times {{16 \times {{10}^{ - 12}}} \over 9}$

= 9 $\times$ 109 $\times$ ${{16 \times {{10}^{ - 12}}} \over 9}$

= 16 mm
3

### 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
4

### 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