Question

Three charges are at the corners of an isosceles triangle. The ±5.00µC charges from a dipole that is 3 cm apart.

1. Find the force (magnitude and direction) the -10.00µC charge, 2 cm from the dipole, exerts on the dipole.
2. For an axis perpendicular to the line connecting the ±5.00µC charges at the mid-point of this line, find the torque (magnitude and direction) exerted on the dipole by the 10.00µC charge.

Answer 1

(1) Electric field due to dipole at equatorial point is given by, $E = \frac{p}{4\pi \epsilon_0 (r^2+a^2)^({3}/{2})}$

where p is the dipole moment, r is the separation of the charge from one charge of the dipole, 'a' is the semi length of the dipole.

Force is given by, $F = qE = \frac{qp}{4\pi \epsilon_0 (r^2+a^2)^({3}/{2})}$

Putting values we get,

$F = qE = \frac{qp}{4\pi \epsilon_0 (r^2+a^2)^({3}/{2})} = \:\frac{9\cdot 10^9\cdot \left(5\cdot 10^{-5}\cdot 0.03\right)\cdot 10\cdot 10^{-6}}{\:\left(0.02^2+0.03^2\right)^{\left(3/2\right)}} = 2880.17 N$ towards negative charge of the dipole parallel to the dipole.

(2) Torque is given by, $\tau = rFsin\theta=pEsin\theta$ where E is the external electric field

and $rsin\theta$ is the perpendicular separation of the two forces.

Here Both forces are intersecting at charge $10\mu C$ . So torque will be zero.