# Electric field due to dipole in polar coordinates

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.

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Orodruin
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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).

Why is there a 1/r term in E along theta? Is sphericcal coordinates beyaond my level?

Orodruin
Staff Emeritus
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