Question

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

• A. \(\frac{-q}{2}\)
• B. \(\frac{-3q}{2}\)
• C. \(\frac{-4q}{3}\)
• D. \(\frac{-4q}{9}\)

Hint:

The middle charge must attract q to cancel the push from +4q, so it is negative. Equate Coulomb forces using distances d/3 and d.

Answer 1

The Correct Option is D

Step 1: Name the two forces
Let $F_Q$ be the force on $q$ due to $Q$ and $F_4$ the force due to $+4q$. Put $q$ at the origin and measure $x$ toward the other charges.

Step 2: Direction check
$F_4$ is repulsive, so it points in the $-x$ direction. For zero net force, $F_Q$ must point in $+x$ toward $Q$. That is attraction, so $Q$ is negative.

Step 3: Use Coulomb's law ratio
Since $F\propto \frac{q_1q_2}{r^2}$, equal forces give $\frac{|Q|}{(d/3)^2}=\frac{4q}{d^2}$.

Step 4: Solve
$|Q|=4q\cdot\frac{(d/3)^2}{d^2}=4q\cdot\frac{1}{9}=\frac{4q}{9}$. So $Q=-\frac{4q}{9}$, option (D).

Final Answer:
Equal and opposite forces need $Q=-\frac{4q}{9}$. \[ \boxed{-\dfrac{4q}{9}} \]