Solved: PHYSICS Question for NEET

NEET PHYSICSMedium

Three charges $+4q$ , $Q$ and $q$ are placed in a straight line of length $l$ at points $0$ , $l/2$ and $l$ distance away from one end respectively. What should be $Q$ in order to make the net force on $q$ to be zero?

$-q$

$4q$

$-q/2$

$-2q$

Step-by-Step Solution

For the net force on charge $q$ (placed at $x=l$ ) to be zero, the force due to $+4q$ (at $x=0$ ) must be equal and opposite to the force due to $Q$ (at $x=l/2$ ).

Force due to $+4q$ : $F_1 = k \frac{(4q)(q)}{l^2}$ (directed away from origin).
Force due to $Q$ : $F_2 = k \frac{(Q)(q)}{(l - l/2)^2} = k \frac{Q q}{(l/2)^2} = k \frac{4Qq}{l^2}$ .

For equilibrium ( $F_{net} = 0$ ): $F_1 + F_2 = 0$ $k \frac{4q^2}{l^2} + k \frac{4Qq}{l^2} = 0$

Dividing by common terms ( $4kq/l^2$ ): $q + Q = 0 \implies Q = -q$ .

(See NCERT Physics Class 12, Chapter 1, Section 1.6).

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