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: Understanding the Question:
This is a problem of electrostatic equilibrium of a charge under the influence of other charges.
Step 2: Key Formula or Approach:
Coulomb's Law: \( F = \frac{k q_1 q_2}{r^2} \).
Net force is the vector sum of forces from all other charges.
Step 3: Detailed Explanation:
Let the positions be \( x_1 = 0 \), \( x_2 = L/2 \), and \( x_3 = L \).
Charges at these positions are \( +3q \), \( Q \), and \( +q \) respectively.
Force on \( +q \) at \( L \) due to \( +3q \) at \( 0 \):
\( F_1 = \frac{k(3q)(q)}{L^2} = \frac{3kq^2}{L^2} \) (directed away from origin).
Force on \( +q \) at \( L \) due to \( Q \) at \( L/2 \):
\( F_2 = \frac{k(Q)(q)}{(L/2)^2} = \frac{4kQq}{L^2} \).
For net force on \( +q \) to be zero:
\[ F_1 + F_2 = 0 \] \[ \frac{3kq^2}{L^2} + \frac{4kQq}{L^2} = 0 \] \[ 3q + 4Q = 0 \implies Q = -\frac{3}{4}q \] Comparing with \( Q = xq \), we get \( x = -3/4 \).
Step 4: Final Answer:
The value of \( x \) is \( -3/4 \).