Question

The electric potential due to an electric dipole at a distance $r$ from the centre of the dipole and at an angle $\theta$ to the dipole axis is ($p$ is the dipole moment, $2a$ is the length of the dipole) :

• A. $V=\frac{p\cos \theta }{4\pi {ϵ}_o(r^2+a^2{\cos }^2\theta )}$
• B. $V=\frac{p\cos \theta }{4\pi {ϵ}_o(r^2-a^2{\cos }^2\theta )}$
• C. $V=\frac{p\cos \theta }{4\pi {ϵ}_o(r^2-a^2{\sin }^2\theta )}$
• D. $V=\frac{p\cos \theta }{4\pi {ϵ}_o(r^2+a^2{\sin }^2\theta )}$

Answer 1

The correct option is B $V=\frac{p\cos \theta }{4\pi {ϵ}_o(r^2-a^2{\cos }^2\theta )}$

From figure, $cos\theta =\frac{r^2+a^2-r_{+}^2}{2ra}$

We get $r_{+}=(r^2+a^2-2ra\cos \theta {)}^{1/2}$

As $r>>a$, we can neglect $a^2$ as compared to $r^2$

$\therefore$ $r_{+}=(r^2-2ra\cos \theta {)}^{1/2}$

Or $r_{+}=r(1-\frac{2a}{r}\cos \theta {)}^{1/2}$

Using $(1+x{)}^n=1+nx$ for $x<<1$

We get $r_{+}=r(1-\frac{a}{r}\cos \theta )$

$⟹$ $r_{+}=r-a\cos \theta$

Similarly, $r_{-}=r+a\cos \theta$

Electric potential at point A $V=V_q+V_{-q}$

$\therefore$ $V=\frac{q}{4\pi {ϵ}_o{r}_{+}}-\frac{q}{4\pi {ϵ}_o{r}_{-}}$

Or $V=\frac{q}{4\pi {ϵ}_o}[\frac{1}{r_{+}}-\frac{1}{r_{-}}]$

Or $V=\frac{q}{4\pi {ϵ}_o}[\frac{1}{r-a\cos \theta }-\frac{1}{r+a\cos \theta }]$

Or $V=\frac{q}{4\pi {ϵ}_o}\left[\frac{2a\cos \theta }{(r-a\cos \theta )(r+a\cos \theta )}\right]$

$⟹$ $V=\frac{p\cos \theta }{4\pi {ϵ}_o(r^2-a^2{\cos }^2\theta )}$