1

### JEE Main 2019 (Online) 10th January Morning Slot

Two electric dipoles, A, B with respective dipole moments ${\overrightarrow d _A} = - 4qai$ and ${\overrightarrow d _B} = - 2qai$ are placed on the x-axis with a separation R, as shown in the figure. The distance from A at which both of them produce the same potential is - A
${{\sqrt 2 R} \over {\sqrt 2 + 1}}$
B
${R \over {\sqrt 2 + 1}}$
C
${{\sqrt 2 R} \over {\sqrt 2 - 1}}$
D
${R \over {\sqrt 2 - 1}}$

## Explanation

V $= {{4qa} \over {\left( {R + x} \right)}} = {{2qa} \over {\left( {{x^2}} \right)}}$

$\sqrt 2 x = R + x$

$x = {R \over {\sqrt 2 - 1}}$ dist $= {R \over {\sqrt 2 - 1}} + R = {{\sqrt 2 R} \over {\sqrt 2 - 1}}$
2

### JEE Main 2019 (Online) 10th January Evening Slot

Charges –q and +q located at A and B, respectively, constitude an electric dipole. Distance AB = 2a, O is the mid point of the dipole and OP is perpendicular to AB. A charge Q is placed at P where OP = y and y >> 2a. The charge Q experiences an electrostatic force F. If Q is now moved along the equatorial line to P' such that OP' = $\left( {{y \over 3}} \right)$, the force on Q will be close to - $\left( {{y \over 3} > > 2a} \right)$ A
9F
B
3F
C
F/3
D
27F

## Explanation

Electric field of equitorial plane of dipole

$= - {{K\overrightarrow P } \over {{r^3}}}$

$\therefore$  At P, F $= - {{K\overrightarrow P } \over {{r^3}}}$Q.

At   P1 , F1 $= - {{K\overrightarrow P Q} \over {{{\left( {r/3} \right)}^3}}} = 27F.$
3

### JEE Main 2019 (Online) 10th January Evening Slot

The Wheatstone bridge shown in figure, here, gets balanced when the carbon resistor used as R1 has the colour code (Orange, Red, Brown). The resistors R2 and R4 are 80$\Omega$ and 40$\Omega$, respectively. Assuming that the colour code for the carbon resistors gives their accurate values, the colour code for the carbon resistor, used as R3, would be - A
Brown, Blue, Brown
B
Grey, Black, Brown
C
Red, Green, Brown
D
Brown, Blue, Black

## Explanation

R1 = 32 $\times$ 10 = 320

for wheat stone bridge

$\Rightarrow$  ${{{R_1}} \over {{R_3}}} = {{{R_2}} \over {{R_4}}}$

${{320} \over {{R_3}}} = {{80} \over {40}}$

${R_3} = 160$

$\therefore$  Correct answer is Brown  Blue  Brown
4

### JEE Main 2019 (Online) 11th January Morning Slot

Three charges Q, + q and + q are placed at the vertices of a right-angle isosceles triangle as shown below. The net electrostatic energy of the configuration is zero, if the value of Q is : A
${{ - q} \over {1 + \sqrt 2 }}$
B
+ q
C
$-$ 2q
D
${{ - \sqrt 2 q} \over {\sqrt 2 + 1}}$

## Explanation U = K$\left[ {{{{q^2}} \over a} + {{Qq} \over a} + {{Qq} \over {a\sqrt 2 }}} \right] = 0$

$\Rightarrow$  q = $-$ Q$\left[ {1 + {1 \over {\sqrt 2 }}} \right]$

$\Rightarrow$  Q = ${{ - q\sqrt 2 } \over {\sqrt 2 + 1}}$