# Electric field calculated by gauss law

• Magnetic Boy
In summary, the conversation discusses the correct interpretation of Gauss's Law for calculating the electric field. The correct answer is option b, which states that the electric field calculated by Gauss's Law is the total field due to all charges, not just the charges enclosed by the surface. This is because Gauss's Law states that the flux of the total electric field through a closed surface is equal to the total charge enclosed, regardless of the other charges present in the universe.
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

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

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

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.

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.

jtbell said:
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.
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.

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]##.

Magnetic Boy
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.

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Magnetic Boy
Delta² said:
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 in the universe...) , 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 universe. 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.
Thanks it helped

## 1. What is the Gauss Law and how is it related to electric fields?

The Gauss Law is a fundamental law in electromagnetism that allows for the calculation of electric fields using the concept of electric flux. It states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space.

## 2. Can the Gauss Law be used to calculate the electric field for any charge distribution?

Yes, the Gauss Law can be used to calculate the electric field for any charge distribution, as long as the charge distribution is symmetric and the electric field is constant over the surface of integration. This is because the Gauss Law assumes a uniform electric field over the surface of integration.

## 3. What is the difference between using Gauss Law and Coulomb's Law to calculate the electric field?

Gauss Law is a more general and easier method for calculating the electric field, as it does not require the use of vector addition and integration. It also works for more complex charge distributions. On the other hand, Coulomb's Law is more specific and only works for point charges or spherically symmetric charge distributions.

## 4. Can the Gauss Law be applied to a charged object in a non-uniform electric field?

Yes, the Gauss Law can still be applied in this scenario, but it would require breaking the charged object into smaller parts and calculating the electric field for each part separately. The total electric field would then be the sum of all the individual electric fields.

## 5. How is the direction of the electric field determined using Gauss Law?

The direction of the electric field can be determined by the direction of the electric flux through the surface of integration. If the electric flux is entering the surface, the electric field is directed towards the enclosed charge. If the electric flux is exiting the surface, the electric field is directed away from the enclosed charge.

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