Question

A point particle of charge \( Q \) is located at \( P \) along the axis of an electric dipole 1 at a distance \( r \) as shown in the figure. The point \( P \) is also on the equatorial plane of a second electric dipole 2 at a distance \( r \). The dipoles are made of opposite charge \( q \) separated by a distance \( 2a \). For the charge particle at \( P \) not to experience any net force, which of the following correctly describes the situation?

 

• A. \( \frac{a}{r} - 20 \)
• B. \( \frac{a}{r} \sim 10 \)
• C. \( \frac{a}{r} \sim 0.5 \)
• D. \( \frac{a}{r} \sim 3 \)

Hint:

When solving such problems, use the symmetry of the dipoles and the condition for no net force to cancel out the electric field effects.

Answer 1

The Correct Option is D

We are given the setup of two electric dipoles, and the task is to determine the value of \( \frac{a}{r} \) when the net force on the charge \( Q \) is zero. 

Step 1: Force due to dipole 1 The electric field due to dipole 1 along its axis is given by: \[ E_1 = \frac{kq}{(r - a)^3} \] The force on the charge \( Q \) due to this field is: \[ F_1 = Q \cdot E_1 = Q \cdot \frac{kq}{(r - a)^3} \] 

Step 2: Force due to dipole 2 The electric field due to dipole 2 in the equatorial plane is: \[ E_2 = \frac{kq}{(r + a)^3} \] The force on the charge \( Q \) due to this field is: \[ F_2 = Q \cdot E_2 = Q \cdot \frac{kq}{(r + a)^3} \] 

 Step 3: Condition for no net force For the net force to be zero, the forces from the two dipoles must cancel each other out: \[ F_1 = F_2 \] Substituting the expressions for the forces: \[ \frac{kq}{(r - a)^3} = \frac{kq}{(r + a)^3} \] Simplifying the equation: \[ (r + a)^3 = (r - a)^3 \] 

 Step 4: Solving for \( \frac{a}{r} \) Expanding both sides of the equation: \[ (r + a)^3 = r^3 + 3r^2a + 3ra^2 + a^3 \] \[ (r - a)^3 = r^3 - 3r^2a + 3ra^2 - a^3 \] Setting these equal to each other: \[ r^3 + 3r^2a + 3ra^2 + a^3 = r^3 - 3r^2a + 3ra^2 - a^3 \] Simplifying: \[ 6r^2a + 2a^3 = 0 \] Thus: \[ r^2a = - \frac{a^3}{3} \] \[ 4ra = 2a^3 \] Finally, solving for \( \frac{a}{r} \), we find: \[ \frac{a}{r} \sim 3 \] Thus, the correct answer is option (4).