Question

Show that how, r1^2 = r^2 + a^2 + 2ar cos(theta)
in Potential due to an electric dipole.

Answer 1

Answer:$r_{1}^{2} = r^{2} +a^{2} + 2ra\ cos\theta$

Explanation: Electric dipole consists of two equal and opposite charges at some distance.

Let us considered two point charges +q and -q.

According to the vector law,

$\vec{a}$ is along to P whereas $\vec{r}$ and $\vec{r}_{1}$ is along to R.

Using vector law of addition

$\vec{r}_{1} = \vec{a} + \vec{r}$

$\theta$ is the angle between direction of $\vec{a}$ and $\vec{r}$.

So, the magnitude of $\vec{r}_{1}$ is

$r_{1}^{2} = r^{2} + a^{2} + 2\vec{r}\cdot\vec{a}$

$r_{1}^{2} = r^{2} +a^{2} + 2ra\ cos\theta$

Hence proved.