Three charges q, Q and +4q are placed in a straight line of length d at points at distance 0, d/3, 2d/3 respectively. In order to make the net force on q be zero, the value of Q should be (a) -q/2 (b) -3q/2 (c) -4q/3 (d) -4q/9

Answer: (b). Force on q from Q (distance d/3): (kqQ)/((d/3)²) = (9kqQ)/(d²). Force on q from 4q (distance 2d/3): (kq(4q))/((2d/3)²) = (9kq²)/(d²). Both act along the same line; for zero net force 9kqQ = -9kq² ⇒ Q = -q. With the data as printed Q = -q, which is not among the options; the key gives (b) -3q/2.

Three charges q, Q and +4q are placed in a straight line of length d… | MHT CET 2026 (11 Apr, Shift 1)

Q14MHT CET 2026 · 11 APR · SHIFT 1Medium

Electric Charge & Coulomb's Law

Three charges $q,\ Q$ and $+4q$ are placed in a straight line of length $d$ at points at distance $0,\ \frac{d}{3},\ \frac{2d}{3}$ respectively. In order to make the net force on $q$ be zero, the value of $Q$ should be

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Answer: (b)

Force on $q$ from $Q$ (distance $d/3$ ): $\dfrac{kqQ}{(d/3)^2} = \dfrac{9kqQ}{d^2}$ .
Force on $q$ from $4q$ (distance $2d/3$ ): $\dfrac{kq(4q)}{(2d/3)^2} = \dfrac{9kq^2}{d^2}$ .
Both act along the same line; for zero net force $9kqQ = -9kq^2 \Rightarrow Q = -q$ .
With the data as printed $Q = -q$ , which is not among the options; the key gives (b) $-\dfrac{3q}{2}$ .

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