Question

electric charge q,q and -2q are placed at the corners of an equilateral triangle of side 'a' find electric dipole moment of the system?​

Answer 1

Answer:

$\large\boxed{\sf{\sqrt{3}qa}}$

Explanation:

Let, there is a ∆ABC in which three point charges +q, +q and -2q are placed on vertices A, B and C respectively.

$\red{Note:-}Refer\;to\;the\; attachment.$

Now, we have to find the net electric dipole moment of the system.

Also, we know that, a dipole is directed from unit negative charge to unit positive charge.

So, let the charge -2q at C , is displaced in two charges of equal magnitude each being equal to -q and -q.

Let, dipole moment $\vec{p_{1}}$ is directed along CA and $\vec{p_{2}}$ along CB.

Also, magnitude of each dipole = p = qa

Here, a is the distance between the charges.

Now, we have two vectors and angle between them is clearly 60°.

Therefore, the magnitude of net electric dipole moment will be equal to,

$= > p_{net} = \sqrt{ {|\vec{p}_{1}|}^{2} + {|\vec{p}_{2}|}^{2} + 2|\vec{p}_{1}||\vec{p}_{2} |\cos60 \degree } \\ \\ = > p_{net} = \sqrt{ {p}^{2} + {p}^{2} + 2 {p}^{2} \times \frac{1}{2} } \\ \\ = > p_{net} = \sqrt{3 {p}^{2} } \\ \\ = > p_{net} = \sqrt{3} p \\ \\ = > \bold{ p_{net} = \sqrt{3} qa}$