Question

Derive the expression for the electric potential at any point along the axial line of an electric dipole?

Answer 1

Please find the solution attached in the image here along

Answer 2

The electric potential is " $\bold{\frac{1}{4 \pi \varepsilon_0}(\frac{2P}{(r^2-4a^2)})}$"

Explanation:

Let P be an axial point r distance from the origin of the dipole. The electrical charge at point P is calculated by the following expression:

$\to V=V_1+V_2$

$V_1$ and $V_2$ charge $+q$ and $-q$ are responsible again for potentials at position P.

$V=\frac{1}{4 \pi \varepsilon_0} (\frac{q}{r-2a} +(\frac{-q}{r+2a}))$

    $=\frac{1}{4 \pi \varepsilon_0} (\frac{4a}{r^2-4a^2})\\\\=\frac{1}{4 \pi \varepsilon_0} (\frac{2P}{r^2-4a^2})$

Learn more:

electric potential: https://brainly.in/question/4580390