# Electric field due to dipole in polar coordinates

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1. Feb 22, 2015

For a dipole, if there is point subtending an angle $\theta$ at the centre of dipole and at a distance $r$ from centre of dipole, then the electric feild at that point can be broken into 2 components. One along the line joining the point and centre of dipole and point given by $$E_r=-\frac{\partial V}{\partial r}$$ and the other component along the perpendicular to that line as $$E_{\theta}=-\frac{1}{r}\frac{\partial V}{\partial \theta}$$
This is given in my book. I understood E_r. But how did they get $E_\theta$ ?
here $V=\frac{pcos\theta}{4\pi\epsilon_0r^2}$ ,r>>a.

2. Feb 23, 2015

### Orodruin

Staff Emeritus
The electric field is minus the gradient of the potential. The equations you have written down are simply the expressions for the components of the gradient in spherical coordinates (apart from the $\varphi$ component that will be zero in this case).

3. Feb 23, 2015

The components of the gradient essentially tell you how much something changes per length in that direction. If you change the radial coordinate by dr, you move a distance dr. Therefore, the radial component is $-\partial V/\partial r$. If you change the coordinate $\theta$ by $d\theta$, then you move a distance $r\,d\theta$ and the $\theta$ component is therefore $-(1/r) \partial V/\partial\theta$.