This question was previously asked in

DSSSB JE EE 2019 Official Paper (Held on 25 Oct 2019)

Option 2 : 0.6 m

__CONCEPT__:

Coulomb’s law: When two charged particles of charges **q1** and **q2** are separated by a distance r from each other then the electrostatic force between them is directly proportional to the multiplication of charges of two particles and inversely proportional to the square of the distance between them.

Force (F) ∝ q1 × q2

\(F \propto \frac{1}{{{r^2}}}\)

\(F = K\frac{{{q_1} \times {q_2}}}{{{r^2}}}\)

Where K is a constant = 9 × 109 Nm2/C2

__Calculations:__

Consider new charge + 4μC is placed d m apart from old +4 μC charge and (x + 0.6) m apart from -16 μC charge.

Let,

q_{A} = + 4 μC at point A

q_{B} = - 16 μC at point B

q_{C} = + 6 μC at point C

Since, net force on charge q_{C} will zero.

∴ |F_{CA}| = |F_{CB}|

From above concept,

\(\frac{Kq_Aq_C}{d^2}=\frac{Kq_Bq_C}{(0.6+d)^2}\)

\(\frac{24}{d^2}=\frac{96}{(0.6+d)^2}\)

4d^{2} = (0.6 + d)^{2}

4d^{2} = 0.36 + d^{2} + 1.2d

3d^{2} - 1.2d - 0.36 = 0

d_{1} = - 0.2 m

d_{2} = + 0.6 m

Hence, according to option **+ 0.6 m distance** should a third charge of + 6 μC be placed from + 4 μC so that no force exerts on it will be zero.