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 determine the electric field at a point \(P\) equidistant from the centers of both dipoles \(A\) and \(B\), we will calculate the electric fields due to each dipole and then find the resultant electric field.

The formula for the electric field due to a dipole at a point along the axial line is given by:

\(E = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p}{r^3}\) 

where \(p = q \times d\) is the dipole moment, \(r\) is the distance from the center of the dipole to the point, and \(\varepsilon_0\) is the permittivity of free space.

For dipole \(A\):

\(q = 2 \, \mu\text{C} = 2 \times 10^{-6} \, \text{C}\)

\(d = 1 \, \text{cm} = 0.01 \, \text{m}\)

\(p_A = q \times d = 2 \times 10^{-6} \times 0.01 = 2 \times 10^{-8} \, \text{Cm}\)

For dipole \(B\):

\(q = 4 \, \mu\text{C} = 4 \times 10^{-6} \, \text{C}\)

\(d = 1 \, \text{cm} = 0.01 \, \text{m}\)

\(p_B = q \times d = 4 \times 10^{-6} \times 0.01 = 4 \times 10^{-8} \, \text{Cm}\)

The distance from each dipole to point \(P\) is half the distance between them:

\(r = \frac{80}{2} = 40 \, \text{cm} = 0.4 \, \text{m}\)

Calculate the electric field due to dipole \(A\):

\(E_A = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p_A}{r^3} = \left(9 \times 10^9\right) \cdot \frac{2 \times 2 \times 10^{-8}}{(0.4)^3}\)

\(E_A = \left(9 \times 10^9\right) \cdot \frac{4 \times 10^{-8}}{0.064} = 5.625 \times 10^4 \, \text{N/C}\)

Calculate the electric field due to dipole \(B\):

\(E_B = \frac{1}{4\pi\varepsilon_0} \cdot \frac{2p_B}{r^3} = \left(9 \times 10^9\right) \cdot \frac{2 \times 4 \times 10^{-8}}{(0.4)^3}\)

\(E_B = \left(9 \times 10^9\right) \cdot \frac{8 \times 10^{-8}}{0.064} = 11.25 \times 10^4 \, \text{N/C}\)

The electric fields \(E_A\) and \(E_B\) are perpendicular to each other. The resultant electric field \(E\) is given by:

\(E = \sqrt{E_A^2 + E_B^2}\)

\(E = \sqrt{(5.625 \times 10^4)^2 + (11.25 \times 10^4)^2} = \sqrt{31.640625 \times 10^8 + 126.5625 \times 10^8}\)

\(E = \sqrt{158.203125 \times 10^8} = \sqrt{1582.03125} \times 10^4 \, \text{N/C} \approx \frac{9}{16}\sqrt{2} \times 10^4 \, \text{N/C}\)

Therefore, the electric field at point \(P\) is \(\frac{9}{16}\sqrt{2} \times 10^4 \, \text{N/C}\).