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1

### JEE Main 2019 (Online) 12th January Morning Slot

Determine the electric dipole moment of the system of the three charges, placed on the vertices of an equilateral triangle, as shown in the figure :

A
$$2q\ell \widehat j$$
B
$$\left( {q\ell } \right){{\widehat i + \widehat j} \over {\sqrt 2 }}$$
C
$$\sqrt 3 \,q\ell {{\widehat j - \widehat i} \over {\sqrt 2 }}$$
D
$$- \sqrt 3 \,q\ell \widehat j$$

## Explanation

$$\left| {{P_1}} \right| =$$ q(d)

$$\left| {{P_2}} \right| =$$ qd

|Resultant| $$=$$ 2 P cos30o

2 qd$$\left( {{{\sqrt 3 } \over 2}} \right)$$ = $$\sqrt 3$$ qd
2

### 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 = $${{ - q\sqrt 2 } \over {\sqrt 2 + 1}}$$
3

### 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 $$= - {{K\overrightarrow P Q} \over {{{\left( {r/3} \right)}^3}}} = 27F.$$
4

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

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