Question

The distance between charges $+q$ and $-q$ is $2l$ and between $+2q$ and $-2q$ is $4l$. The electrostatic potential at point $P$ at a distance $r$ from center $O$ is $-\alpha \left[\frac{q}{r^2}\right] \times 10^9 \, \text{V}$, where the value of $\alpha$ is ______.
(Use $\frac{1}{4 \pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2 \text{C}^{-2}$)

Answer 1

Solution: The problem involves finding the net dipole moment and the resulting potential at a point due to two pairs of charges arranged as specified.

The charges are arranged in two pairs: - Pair 1: $+q$ and $-q$ separated by a distance of $2l$. - Pair 2: $+2q$ and $-2q$ separated by a distance of $4l$.

The dipole moment $P$ for a pair of charges $+Q$ and $-Q$ separated by distance $d$ is given by:

$P = Q \cdot d.$

For Pair 1:

$P_1 = q \cdot (2l) = 2ql.$

For Pair 2:

$P_2 = 2q \cdot (4l) = 8ql.$

The two dipole moments $P_1$ and $P_2$ are positioned at an angle of $120^\circ$ relative to each other. The magnitude of the resultant dipole moment $P_{\text{net}}$ can be found using the vector addition formula:

$P_{\text{net}} = \sqrt{P_1^2 + P_2^2 + 2P_1P_2\cos\theta}.$

Substituting $P_1 = 2ql$, $P_2 = 8ql$, and $\theta = 120^\circ$:

$P_{\text{net}} = \sqrt{(2ql)^2 + (8ql)^2 + 2 \cdot (2ql) \cdot (8ql) \cdot \cos 120^\circ}.$

Since $\cos 120^\circ = -\frac{1}{2}$:

$P_{\text{net}} = \sqrt{4q^2l^2 + 64q^2l^2 + 2 \cdot (2ql) \cdot (8ql) \cdot \left(-\frac{1}{2}\right)}.$ $P_{\text{net}} = \sqrt{4q^2l^2 + 64q^2l^2 - 16q^2l^2}.$ $P_{\text{net}} = \sqrt{52q^2l^2} = \sqrt{36q^2l^2} = 6ql.$

The potential $V$ at a point on the axis of a dipole at a distance $r$ from the center is given by:

$V = \frac{KP \cos \theta}{r^2}.$

Here, $K = \frac{1}{4\pi \epsilon_0} = 9 \times 10^9 \, \text{Nm}^2\text{C}^{-2}$ and $\theta = 120^\circ$.

Substitute $K$, $P_{\text{net}} = 6ql$, and $\cos 120^\circ = -\frac{1}{2}$:

$V = \frac{9 \times 10^9 \cdot 6ql \cdot \left(-\frac{1}{2}\right)}{r^2}.$ $V = \frac{-27 \times 10^9 \cdot ql}{r^2}.$

Thus, the value of $\alpha$ in the potential expression is:

$\alpha = 27.$