Question

Derive an expression for the electric potential at a point along the axial line of an electric dipole. At a point charge, the values of electric field and potential are $25N/C$ and $10J/C$. Then calculate
(a) magnitude of the charge and
(b) distance of the charge from the point of observation

Answer 1

Let $P$ be the axial point at distance $x$ from the centre of the dipole electric potential $V$ at point $P$ is given by,

$V=V_{+q}+V_{-q}$

$=K\left\{\frac{q}{x-a}\frac{-q}{x+a}\right\}$

$=\frac{Kq2a}{x^2-a^2}=\frac{KP}{(x^2-a^2)}$

$P:$ dipole moment $=2aq$

${II}^{nd}$ Part

2) $E=\frac{V}{r}\Rightarrow 25=\frac{10}{r}$

$r=\frac{10}{25}=0.4m$

$\therefore$ distance of the charge from point of observation $=0.4m$

Now, i) $V=\frac{KQ}{r}$

$\Rightarrow 10\times 0.4=9\times {10}^9\times Q$

$\therefore Q=4.45\times {10}^{-10}C$

$\therefore$ Magnitude of charge $=4.45\times {10}^{-10}C$

distance $=0.4m$