Question

Three charges $4q,Q$ and $q$ are in a straight line in the position of 0, $l/2$ and $l$ respectively. The resultant force on $q$ will be zero, if $Q =$

• A. – q
• B. $- 2q$
• C. $- \frac{q}{2}$
• D. $4q$

Answer 1

Answer: (A)

– q

Answer 2

The force between 4q and q; $F_{1} = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{4q \times q}{l^{2}}$

The force between Q and q;

$F_{2} = \frac{1}{4\pi\varepsilon_{0}} \cdot \frac{Q \times q}{(l/2)^{2}}$

We want $F_{1} + F_{2} = 0$ or $\frac{4q^{2}}{l^{2}} = - \frac{4Qq}{l^{2}}$ ⇒ $Q = - q$