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I know we can find out the electric field using the electric field $$E=\frac{KQ}{R^2}$$ taking small element $dq$ and finding the electric field by integrating the value of $dE$ over the circumference which will be $$E=\frac{kxQ}{\sqrt {(a^2+x^2)^3}}$$ where $a$ is radius of ring and $x$ is distance of point $p$ on the axis of the ring. Can we find the same using the Gauss law?

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  • $\begingroup$ How would you use Gauss' Law? $\endgroup$ – sammy gerbil Jun 7 '17 at 15:01
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Gauss' law uses symmetry for its functioning.

Even if you consider a cylindrical gaussian (or spherical) surface that encloses the ring, the electric field lines will be very different for two adjacent dA area elements and you will have to consider each and every of these $dA$ area elements of the gaussian surface to get the electric field. Thats certainly not how integration is supposed to work. It would make the problem harder to solve.

This integration is like adding $1+2+3+4+5+6=21$, i.e, you will have to consider everything. That's impossible to do.

But if the electric field lines and their dot product with corresponding $dA$ was same for every field line then integration would become simpler. Such is the case when electric field due to infinite sheet is to be calculated.

This integration of this type is like adding $2+2+2+2+2+2=6\times2$. Now this is simpler as it assumes symmetry.

enter image description here

(The image shows electric field due to 30 charges arranged in a ring at a given observation point. The position of the observation point can be varied to see how the electric field of the individual charges adds up to give the total field.)

In short, the electric field lines due to the ring lack symmetry. This would only make the problem harder to solve using Gauss' law.

$\oint E.dA=\frac{q}{\epsilon_{\circ}} $

Recall when the electric field due to a infinite sheet is to be found, a cylindrical gaussian surface is considered that encloses a circular part of the sheet inside it. The electric field lines for this sheet are perpendicular to it. This is a perfectly ideal condition for Gauss' law to work. The field lines emerge perpendicularly from the 2 circular parts of the cylinder.

enter image description here

But in case of a ring, such conditions are absent.

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