# I Electric field calculated by gauss law

1. Jul 8, 2016

### Magnetic Boy

Which one is correct??

1) electric field calculated by gauss law is the field due to charges inside the closed surface.

2) the flux of the electric field through a closed surface due to all charges is equal to the flux due to charges enclosed by the surface

2. Jul 9, 2016

### hackhard

both
ist one is true only for gaussian surfaces with symmetric charge distribution inside

3. Jul 9, 2016

### Magnetic Boy

This is a single answer question. Which one is more appropriate? Why?

4. Jul 9, 2016

### Staff: Mentor

Tell us what you think the answer is, and why. Then we can tell you whether your answer and reasoning are correct, and if they are not, steer you in the right direction.

5. Jul 9, 2016

### vanhees71

Hint: Write down the equation and think carefully about the definitions of the involved surface and volume integral. Also note that in Gauß's Law the total $\vec{D}$ field and the charge distribution enters.

Of course, the Maxwell equations in differential form are the natural way to think about electromagnetism, which is the paradigmatic example of a local relativistic field theory and as such is much easier interpreted in terms of the partial differential equations of motion of the fields and its sources (charge-current density) than in integral form.

6. Jul 9, 2016

### Magnetic Boy

I think option b is correct. I think electric field that we find from guass law is the field of not only enclosed charges. For example if i have a charged stick of length 1 meter and i draw a guassian surface which encloses small part of the stick. Now if i apply guass law on that surface and get the electric field. That electric field will be equal to the electric field of the whole stick.. but i m not sure. Thats why i posted.

7. Jul 9, 2016

### vanhees71

Yes, it's correct. Gauß's Law in integral form says (in Heaviside-Lorentz units)
$$\int_{\partial V} \mathrm{d} \vec{F} \cdot \vec{E}=\int_V \mathrm{d}^3 \vec{x} \rho(\vec{x}).$$
Here $V$ is some volume and $\partial V$ the boundary surface. By definition, the boundary-surface elements are directed perpendicular outward of the considered volume.

Try it with your example. Hint: The electrostatic potential in this case is given by
$$\Phi(\rho,z)=\int_{-L/2}^{L/2} \mathrm{d} z' \frac{\lambda}{4 \pi \sqrt{\rho^2+(z-z')^2}}, \quad \rho=\sqrt{x^2+y^2}.$$
Now calulate the electric field and integrate it over a cylinder around the $z$ axis that encloses only a part of the line charge, which here I assumed to be on the $z$ axis in the interval $z \in [-L/2,L/2]$.

8. Jul 9, 2016

### Delta²

Interesting choices, but we can see easily that 1) is not correct, if we have study carefully what Gauss's law says. The electric field in Gauss's law is the total electric field (simply put from all the charges inside and outside the closed surface...) , Gauss's law just tell us that the flux of the (total) electric field through a closed surface equals the total charge enclosed by the surface (divided by a constant). So in calculating that flux we take into account only the charges enclosed, regardless of the charges that exist in the rest of the space. So in calculating that flux is like only the enclosed charges are present, so it is equal to the flux due to the enclosed charges.

Last edited: Jul 9, 2016
9. Jul 9, 2016

### Magnetic Boy

Thanks it helped