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

To find the electric field at point \( P \), equidistant from the centers of both dipoles, we must calculate the electric field contributions from both dipoles and combine them. The formula for the electric field due to a dipole at a point along the perpendicular bisector is given by:

\(E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{p}{r^3}\) 

where:

• \(p\) is the dipole moment, given by \(p = q \times d\),
• \(r\) is the distance from the center of the dipole to the point where the field is being calculated,
• \(\varepsilon_0\\) is the permittivity of free space \((8.85 \times 10^{-12}\ \text{C}^2/\text{N}\cdot\text{m}^2)\).

Calculating the Dipole Moments:

• Dipole \( A \): \( q = 2\ \mu\text{C} = 2 \times 10^{-6}\ \text{C}\), \(d = 1\ \text{cm} = 0.01\ \text{m}\)
• Dipole Moment \( p_A = q \times d = 2 \times 10^{-6} \times 0.01 = 2 \times 10^{-8}\ \text{C}\cdot\text{m}\)
• Dipole \( B \): \( q = 4\ \mu\text{C} = 4 \times 10^{-6}\ \text{C}\), \(d = 1\ \text{cm} = 0.01\ \text{m}\)
• Dipole Moment \( p_B = q \times d = 4 \times 10^{-6} \times 0.01 = 4 \times 10^{-8}\ \text{C}\cdot\text{m}\)

Calculating Electric Fields at Point \( P \):

• Distance \( r = \sqrt{(40^2 + 40^2)} = 40\sqrt{2}\ \text{cm} = 0.4\sqrt{2}\ \text{m}\)
• Electric field due to dipole \( A \):
  \(E_A = \frac{1}{4\pi \varepsilon_0} \cdot \frac{p_A}{r^3} = \frac{9 \times 10^9 \times 2 \times 10^{-8}}{(0.4\sqrt{2})^3} = \frac{9 \times 10^9 \times 2 \times 10^{-8}}{0.064\sqrt{8}}\ \text{N/C}\)
• Electric field due to dipole \( B \):
  \(E_B = \frac{1}{4\pi \varepsilon_0} \cdot \frac{p_B}{r^3} = \frac{9 \times 10^9 \times 4 \times 10^{-8}}{(0.4\sqrt{2})^3} = \frac{9 \times 10^9 \times 4 \times 10^{-8}}{0.064\sqrt{8}}\ \text{N/C}\)

Resultant Electric Field:

Since both \( E_A \) and \( E_B \) are in the perpendicular direction, the resultant field is:

\(E = E_B - E_A = \frac{9 \times 10^4}{\sqrt{2}}\sqrt{2} = 9\times 10^4\ \text{N/C}\)

Conclusion:

The correct answer is \(\frac{9}{16}\sqrt{2} \times 10^4\) N/C, thus eliminating the other options.

Diagram showing dipoles \( A \) and \( B \) and point \( P \).