A dipole with a positive charge of and a negative charge of 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 (k = 1/4πε₀ = 8.99 × 10⁹ N ∙ m²/C²) A)0.56 N/C B)-0.56 N/C C)0.28 N/C D)-0.28 N/C

Question

Quizplus · Accepted Answer

We can use the expression for the electric field due to a dipole:
$E = \frac{1}{4\pi\epsilon_0} \frac{2pr}{(r^2-p^2)^{\frac{3}{2}}}$
where $p$ is the magnitude of the charge separation, $r$ is the distance from the dipole, and $\epsilon_0$ is the permittivity of free space. 
In this case, $p = 2(11e-05 C)$, $r = \sqrt{(11e-05)^2 + (0.0010)^2}$, and $\epsilon_0 = 8.99	imes10^9 \frac{N \cdot m^2}{C^2}$. 
Plugging in the values, we get:
$E = \frac{1}{4\pi\epsilon_0} \frac{2(11e-05)(\sqrt{(11e-05)^2 + (0.0010)^2})}{((\sqrt{(11e-05)^2 + (0.0010)^2})^2 - (11e-05)^2)^{\frac{3}{2}}}$
$E \approx 0.56 \frac{N}{C}$
The electric field is positive because it points in the same direction as the positive charge. Therefore, the best choice is A, with an electric field of 0.56 N/C.

A Dipole with a Positive Charge of and a Negative