Question

Two short dipoles \( (A, B) \), \( A \) having charges \( \pm 2\,\mu\text{C} \) and length \( 1\,\text{cm} \) and \( B \) having charges \( \pm 4\,\mu\text{C} \) and length \( 1\,\text{cm} \) are placed with their centres \( 80\,\text{cm} \) apart as shown in the figure. The electric field at a point \( P \), equidistant from the centres of both dipoles is \underline{\hspace{2cm}} N/C.

• A. $4.5\sqrt{2} \times 10^4$
• B. $9\sqrt{2} \times 10^4$
• C. $\frac{9}{16}\sqrt{2} \times 10^5$
• D. $\frac{9}{16}\sqrt{2} \times 10^4$

Hint:

Axial field is twice the equatorial field for same distance and dipole moment.

Answer 1

The Correct Option is D

$r = 0.4$ m. $p_A = 2 \times 10^{-8}$. $p_B = 4 \times 10^{-8}$.
$E_A$ (Axial) $= 2kp_A/r^3 = 4k \times 10^{-8}/r^3$.
$E_B$ (Equatorial) $= kp_B/r^3 = 4k \times 10^{-8}/r^3$.
$E_{net} = E \sqrt{2}$.
$E = \frac{9 \times 10^9 \times 4 \times 10^{-8}}{0.064} = 5625$.
$E_{net} = 5625\sqrt{2} = \frac{90000}{16}\sqrt{2} = \frac{9}{16}\sqrt{2} \times 10^4$.