A dipole with a positive charge of 2.0 μC and a negative charge of -2.0 μC is centered at the origin and oriented along the x-axis with the positive charge located to the right of the origin. The charge separation is 0.0010 m. Find the electric field due to this dipole at the point x = 4.0 m, y = 0.0 m. (k = 1/4πε0 = 8.99 × 10⁹ N ∙ m²/C²) A) 0.56 î N/C B) -0.56 î N/C C) 0.28 î N/C D) -0.28 î N/C

Question

Quizplus · Accepted Answer

Using the equation for the electric field due to a point charge:

$E=\frac{kq}{r^2}$

where k is the Coulomb's constant, q is the magnitude of the charge, and r is the distance between the charge and the point where the electric field is being calculated.

For a dipole, the electric field at a point P along the axial line of the dipole is given by:

$E= \frac{k p}{r^3}$ $(+)$ or $(-) $ ,$ \vec{a}$

where $p$ is the dipole moment, $r$ is the distance from the dipole to the point P, and $\vec{a}$ is the unit vector along the axial line and pointing towards the dipole's positive charge.

In this problem, the magnitude of the dipole moment is given by:

$p=qd=2.0 × 10^{-6} C \cdot 0.0010 m=2.0 × 10^{-9} C \cdot m$

The distance from the dipole to the point P is:

$r=\sqrt{(4.0 m)^2+(0.0 m)^2}=4.0 m$

The unit vector along the axial line is:

$\vec{a}=\frac{+4.0 \hat{i}}{4.0}=+ \hat{i}$

(we choose the positive x-axis direction as the direction for $\vec{a}$ since the positive charge is to the right of the origin)

Substituting these values into the equation for the electric field, we get:

$E= \frac{(8.99 ×10^9 N \cdot m^2/C^2)(2.0 × 10^{-9} C \cdot m)}{(4.0 m)^3} \hat{i}$

$E=0.56 \hat{i} N/C$

Therefore, the electric field due to the dipole at the point P is 0.56 î N/C, which is answer choice A.

A Dipole with a Positive Charge of 2