JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Electrostatics

Four charges

Q_{1}, Q_{2}, Q_{3}

and

Q_{4}

of same magnitude are fixed along the

x

axis at

x = - 2 a, - a, + a

and

+ 2 a

, respectively. A positive charge

q

is placed on the positive

y

axis at a distance

b > 0

. Four options of the signs of these charges are given in List I. The direction of the forces on the charge

q

is given in List II. Match List I with List II and select the correct answer using the code given below the lists.

Answer: (A)

Solution

P: All charges positive → all repel

q

. By symmetry,

x

-components cancel, only

+ y

component remains. P→3.

Q:

Q_{1}, Q_{2}

positive (left side),

Q_{3}, Q_{4}

negative (right side) → left charges repel, right charges attract. Net force is toward

+ x

. Q→1.

R:

Q_{1}, Q_{4}

positive (outer),

Q_{2}, Q_{3}

negative (inner) → outer charges repel (weaker, farther), inner charges attract (stronger, closer). The

y

-component of attraction from inner charges dominates, giving net force in

- y

. R→4.

S:

Q_{1}, Q_{3}

positive (at

- 2 a, + a

Q_{2}, Q_{4}

negative (at

- a, + 2 a

) → forms two dipoles. The

Q_{2}

Q_{3}

dipole (closer) has stronger

- x

component than the

Q_{1}

Q_{4}

dipole's

+ x

component. Net force is

- x

. S→2.

Answer: P-3, Q-1, R-4, S-2.

JEE Advanced · 2014 · 25 May 2014 · Paper 2 · Single correct option