Question

Three charges \( +3q, Q \) and \( +q \) are placed in a straight line of length \( L \) at points at distances \( 0, \frac{L}{2} \) and \( L \) respectively. The value of \( Q \) in order to have the net force on \( +q \) to be zero, \( Q = xq \). The value of \( x \) is

• A. \( \frac{1}{4} \)
• B. \( -\frac{3}{4} \)
• C. -3
• D. 4

Hint:

To balance two positive charges, a central charge must be negative.

Answer 1

The Correct Option is B

Step 1: Equilibrium Condition
Net force on charge $+q$ (at $x=L$) must be zero: $F_{3q} + F_Q = 0$.
Step 2: Force Equations
$\frac{k(3q)(q)}{L^2} + \frac{k(Q)(q)}{(L/2)^2} = 0$
Step 3: Solving for Q
$\frac{3q}{L^2} + \frac{4Q}{L^2} = 0 \Rightarrow 4Q = -3q \Rightarrow Q = -\frac{3}{4}q$
Step 4: Finding x
Since $Q = xq$, then $x = -3/4$.
Final Answer:(B)