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CASE 1: Consider an enclosed spherical surface with a charge $q$ at its centre. From Gauss' law we can say that the flux through this sphere is $q/\epsilon_0$.

CASE 2: The charge is inside but off centre. In this case the electric field will not be normal to the surface. So how can we use the formula $q/\epsilon_0$? It turned out that the answer is same for both cases. This is my confusion.

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calculate it. it will come that only. – evil999man Apr 15 '14 at 17:23
but the doubt is how can we use the abv formula when the field is not normal to the surface?? – vas Apr 15 '14 at 17:25
the formula is valid. you have to use calculus to calculate it as r and theta is changing – evil999man Apr 15 '14 at 17:32
If you like this question you may also enjoy reading this Phys.SE post. – Qmechanic Apr 15 '14 at 19:12
up vote 4 down vote accepted

You can use $q/\epsilon_0$ to calculate flux for both cases because that's what Gauss' law says: Just look at the enclosed charge. It's amazing.

Flux for any closed surface is $\Phi = \oint \vec{E} \cdot d\vec{A}$. There are two ways to calculate this quantity:

  1. The hard way, which means evaluating $\vec{E}$ on every part of the surface, and integrating. This is hard because for your case 2, the electric field strength varies on the surface and it is not normal everywhere. When computed in this manner, it is not obvious at all that your two cases should yield the same value for flux. It really isn't obvious.
  2. The easy way, which is using Gauss' law $\Phi = Q_\text{encl}/\epsilon_0$. To evaluate flux using Gauss' law, you only need to look at how much charge how is physically enclosed. That's what Gauss' law says. That's why Gauss' law gives the same answer in both cases. It makes no mention of the particular geometry of your closed surface and does not assume that the electric field is normal to the surface.

There are proofs of Gauss' law floating around, but I don't think you are asking for that.

As a side note, one often uses Gauss' law to figure out what the electric field is. Case 1 is very useful for figuring out the electric field at some distance away from your charge. Case 2 is not useful since you can't factor out $\vec{E}$ from the integral. Nonetheless, both cases yield the same value for flux.

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does that mean we can get differant answers by calculating from above two methods you suggest?? – vas Apr 15 '14 at 17:35
No, both methods for calculating flux will field the same value. Gauss' law just gives a shortcut to calculating it. – BMS Apr 15 '14 at 17:38
but dont we derive gauss law assuming that field lines are normal to the surface?? – vas Apr 15 '14 at 17:39
No. A proper mathematical proof of Gauss' law will show that no such assumption is made. But Gauss' law is usually only used is such cases, which might be leading to your confusion. – BMS Apr 15 '14 at 17:41
alright.....thankyou so mucchh – vas Apr 15 '14 at 17:42

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