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 centre $$O$$ is $$-\alpha\frac{ql}{r^2} \times 10^9$$ V, where the value of $$\alpha$$ is. (Use $$\frac{1}{4\pi\varepsilon_0} = 9 \times 10^9$$ N m$$^2$$ C$$^{-2}$$)

step 1: positions of charges

On the x-axis:

• +2q at −2l
• −q at −l
• +q at +l
• −2q at +2l

step 2: net charge

$$2q-q+q-2q=0$$

So monopole term = 0

step 3: dipole moment

Take dipole moment about O:

$$p=\sum_{ }^{ }q_ix_i$$

$$=(2q)(-2l)+(-q)(-l)+(q)(l)+(-2q)(2l)$$

So,

$$p=-6ql$$

(direction along negative x-axis)

step 4: potential at point P

For dipole:

$$V=\frac{1}{4\pi\epsilon_0}\frac{p\cos\theta}{r^2}$$

Here $$\theta=60^{\circ}$$

$$\cos60^{\circ}=\frac{1}{2}$$

$$V=9\times10^9\cdot\frac{-6ql\cdot(1/2)}{r^2}$$

$$=9\times10^9\cdot\frac{-3ql}{r^2}$$