Question

Net electric field at point A as shown in figure is at an angle of $60^\circ$ with x-axis then, find $\frac{P_2}{P_1} = ?$ 

• A. $\frac{1}{\sqrt{3}}$
• B. $2\sqrt{3}$
• C. $\sqrt{3}$
• D. $\sqrt{3}/2$

Answer 1

The Correct Option is B

Step 1: Calculate electric field components at point A due to dipole $P_1$ (axial) and $P_2$ (equatorial).
$E_1 = \frac{2kP_1}{r^3}$ (axial field, along x-axis).
$E_2 = \frac{kP_2}{r^3}$ (equatorial field, along y-axis).

Step 2: Use the given direction of the net electric field.
$\tan 60^\circ = \frac{E_2}{E_1}$.

Step 3: Substitute the expressions and solve for the ratio.
$\sqrt{3} = \frac{kP_2/r^3}{2kP_1/r^3} = \frac{P_2}{2P_1}$.
$\frac{P_2}{P_1} = 2\sqrt{3}$.

Final Answer: Option (2).