If we take a Gauss surface S to calculate the electric field using Gauss law: $$\int\epsilon\vec{E}.\vec{dS}=\sum(Q's\ inside\ S\ + Q's\ on\ S\over2 )$$

then this field calculated $E$ is for what?Is this $E$ for each point(individually) on Gauss surface taken only i.e. do we mean by this E the electric field done due to the charges covered by the Gauss surface taken,done on each unique point on Gauss surface chosen?or is it also for points inside the surface?(short answer)

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    – DanielSank
    Commented May 18, 2015 at 0:45

1 Answer 1


Any formula used in physics is for calculating exactly what it says. The formula above says :

If you want to know the integrated electric field on a surface sum the quantities on the right. So it will yield one real number, in units of charge.

  • $\begingroup$ ok this didn't answer me. my question was this E that will be calcullated from this formula, is it the electric field of a point on Guass surface taken?And if yes,and if we have a charge q outside the guass surface so to calculate E on a point on Guass surface taken,wwe should add E calculated by Guass on this point to the E due to q right?(please answer both questions) $\endgroup$
    – HHH
    Commented May 18, 2015 at 5:49
  • $\begingroup$ that is the misunderstanding: "E that will be calcullated from this formula" E is not calculated, an integral of E is calculated on the surface and the formula says that it will look like a charge Q +q, i.e. the result of the integration, which is the sum of the charges on the surface and inside. To get the electric field you need the differential form of Gauss's law and will have to solve a differential equation and apply the boundary conditions. $\endgroup$
    – anna v
    Commented May 18, 2015 at 7:18

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