0
$\begingroup$

We have 3 point charges $Q_1, Q_2$ and $Q_3$ and know electric fields $E_1, E_2$ and $E_3$ ($E_1$ - field due to $Q_1$ etc., $E_1+E_2+E_3 = E$ is electric field in any point of space), and we enclose $Q_1$ and $Q_2$ with some surface, then if we wanted to get $\frac{Q_1+Q_2}{\epsilon_0}$ (right side of Gauss law) what field should sum over surface (left side of Gauss law)? $E_1+E_2$ (field due to enclosed charges) or $E$(the electric field)?

In other words is $E$ in Gauss law ( $ \oint_S\vec E d\vec A = \frac{Q}{\epsilon_0} $) field due to all charges or only ones enclosed in some surface S?

$\endgroup$
1
$\begingroup$

If $Q_3$ isn't enclosed by the sphere of interest, then the net contribution of $E_3$ to the integral $\oint_S E\ dA$ is going to be $0$.

That is, $$\oint_S E\ dA = \oint_S (E_1 + E_2 + E_3)\ dA$$ $$ = \oint_S (E_1 + E_2)\ dA + \oint_S E_3\ dA$$

and since,

$$\oint_S E_3\ dA = 0$$ we have,

$$\oint_S E\ dA = \oint_S (E_1 + E_2)\ dA $$

Thus you should arrive at the same answer for the surface integral whether or not the contribution of $Q_3$ to the field is included in the integral or not.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.