# Electric field on Gaussian surface due to external charge

• NoahCygnus
In summary, the Gauss' law does not give me the net electric field at point P because the flux through the sphere is zero due to the charge +Q. However, you can still calculate the electric field due to the charge +q by simply superposing the electric fields generated by the charge +Q and the charge itself.
NoahCygnus

There's something I need to confirm about Gauss' law. If I have to determine the electric field at point P due to charge +q, I take a Gaussian sphere enclosing the charge with the point on the surface of the sphere. So Gauss law doesn't care about the charge +Q because the flux do to this charge is zero. But clearly the electric field of +Q will also influence the net E field at point P according to superposition principle, hello E field is a vector and the E fields of Q and q will add according to vector laws at point p. So I conclude that the Gauss' law doesn't actually give me the net electric field at point P , but only the electric field due to the charge +q, am I correct?

#### Attachments

• TaTFo5H.jpg
11.2 KB · Views: 1,639
NoahCygnus said:
So I conclude that the Gauss' law doesn't actually give me the net electric field at point P , but only the electric field due to the charge +q, am I correct?
No. Gauss' Law says that you can deduce how much charge is enclosed by the Gaussian sphere from the net electric flux through the sphere. The charge inside the sphere creates a net flux, but the charge outside does not. Gauss's Law allows you to calculate the electric field in a few cases of high symmetry. This isn't one of them.

kuruman said:
No. Gauss' Law says that you can deduce how much charge is enclosed by the Gaussian sphere from the net electric flux through the sphere. The charge inside the sphere creates a net flux, but the charge outside does not. Gauss's Law allows you to calculate the electric field in a few cases of high symmetry. This isn't one of them.
In MIT 8.02 , the professor ignored the charge outside the Gaussian sphere when he calculated E fields. I don't get it, how is the above situation not symmetric? There is a nice sphere enclosing a point charge located at its centre.

And there is also a second, not so nice, charge that breaks the spherical symmetry. When you use Gauss' Law to find the E-field, you must first ascertain that (a) the field is perpendicular to the surface at all its points and (b) the field has the same magnitude at all points on the surface. You need (a) so the you can say that ##\vec E \cdot \hat n = E## and you need (b) to say that ##E## is constant and take it out of the surface integral. Both (a) and (b) are not the case here. However, you can find E at point P and all other point on the sphere by simple superposition of the electric fields generated by the two charges.

If we write, as Kuruman says,
Etot = Eq + EQ
as the total field due to both charges (indeed the superposiion of any other charges in the universe), and then find the flux through the spherical surface, of this total field, it will, according to Gauss' law, be equal to the charge inside the sphere (which, in this case, is q) divided by ε0

kuruman said:
And there is also a second, not so nice, charge that breaks the spherical symmetry. When you use Gauss' Law to find the E-field, you must first ascertain that (a) the field is perpendicular to the surface at all its points and (b) the field has the same magnitude at all points on the surface. You need (a) so the you can say that ##\vec E \cdot \hat n = E## and you need (b) to say that ##E## is constant and take it out of the surface integral. Both (a) and (b) are not the case here. However, you can find E at point P and all other point on the sphere by simple superposition of the electric fields generated by the two charges.
I apologize for the late reply. I think I get it now. But I got a question, as in the above situation, does ##\vec{E}## in the equation ##\oint \vec{E}.d\vec{A}=q/\epsilon_{0}## represent total electric field due to both charges outside and inside or electric field only due to charge +q? I think it represents e field due to both charges +q and +Q but as ##\oint [\vec{E_{+q}} +\vec{E_{+Q}}].d\vec{A} = q/\epsilon_{0} \Rightarrow \oint \vec{E_{+q}}.d\vec{A}+\oint \vec{E_{+Q}}.d\vec{A} = q/\epsilon_{0}##The integral ##\oint\vec{E_{+Q}}.d\vec{A}=0##, the equation becomes ##\oint \vec{E_{+q}}.d\vec{A}=q/\epsilon_{0}## so we only consider e field due to charge +q in the Gauss' equation. Right?

Right. The surface integral of the electric field due to ##+Q## alone is zero. The intuitive way to think about it is this. Whatever electric filed lines are generated by ##+Q## that enter the closed surface from one side must come out of the other side. The same is not true for ##+q## inside. Whatever lines are generated by it, just cross the surface.

kuruman said:
Whatever electric filed lines are generated by +Q that enter the closed surface from one side must come out of the other side.
This isn't necessarily true in general. The pattern of field lines due to two charges isn't simply the union of the field lines produced by each one alone. Suppose the charge outside the Gaussian surface is + and the one inside is -. Then some or even all of the field lines from the + charge enter the surface but never leave, because they end at the - charge.

Nevertheless, the number of lines leaving minus the number of lines entering is proportional to the charge inside. In this case it would be negative (more lines entering than leaving) because the charge inside is -.

jtbell said:
This isn't necessarily true in general. The pattern of field lines due to two charges isn't simply the union of the field lines produced by each one alone.
I agree, it is the superposition of two contributions at each point. However, one can always draw radial electric field lines for each individual point charge and look at the resulting picture in order to calculate the flux. In other words, instead of adding vectorially the two electric fields point by point and then calculating the flux, one can calculate the flux over the spherical surface from each charge separately and then add fluxes. Superposition allows this to be done.

Let ##q_{in}## be the charge enclosed by the Gaussian sphere and ##q_{out}## be the charge outside the Gaussian sphere. Also, let ##\Phi_{in}##, ##\vec{E}_{in}## and ##\Phi_{out}##, ##\vec{E}_{out}## be the electric flux, electric field due to each charge alone. Then net flux through the Gaussian surface is
$$\Phi_{net}=\oint \vec{E}_{net}.d\vec{A}=\oint (\vec{E}_{out}+\vec{E}_{in}).d\vec{A}=\oint \vec{E}_{out}.d\vec{A}+\oint \vec{E}_{in}.d\vec{A}=\Phi_{out}+\Phi_{in}$$ The integration is over the Gaussian sphere that excludes ##q_{out}##. One can now argue that the pattern of radial lines formed by ##q_{out}## shows the same number of lines going in the Gaussian sphere as going out, i.e. ##\Phi_{out}=0##. (More rigorously the argument involves the solid angle ##d\Omega## subtended by the areas ##dA## of entry and exit and is found in E&M textbooks.) In any case, the first integral is zero, therefore ##\Phi_{net}=\Phi_{in}=\oint \vec{E}_{in}.d\vec{A}##.

## 1. What is an electric field on a Gaussian surface?

An electric field on a Gaussian surface is the measure of the strength and direction of the electric field at any point on the surface. It is a vector quantity, meaning it has both magnitude and direction.

## 2. How is the electric field on a Gaussian surface affected by an external charge?

The electric field on a Gaussian surface is affected by an external charge through the principle of superposition. This means that the electric field caused by the external charge is added to the existing electric field at each point on the surface.

## 3. What is the relationship between the electric field and the Gaussian surface?

The electric field on a Gaussian surface is perpendicular to the surface at every point. This means that the electric field lines are always perpendicular to the surface and do not intersect it.

## 4. Can the electric field on a Gaussian surface be negative?

Yes, the electric field on a Gaussian surface can be negative. A negative electric field indicates that the direction of the field is towards the surface, while a positive electric field is directed away from the surface.

## 5. How is the electric field on a Gaussian surface calculated?

The electric field on a Gaussian surface can be calculated using Gauss's law, which states that the electric flux through a closed surface is equal to the total charge enclosed within the surface divided by the permittivity of the medium. This can be expressed as E = Q/ε0, where E is the electric field, Q is the charge enclosed, and ε0 is the permittivity of free space.

• Electromagnetism
Replies
83
Views
3K
• Electromagnetism
Replies
3
Views
804
• Electromagnetism
Replies
2
Views
169
• Electromagnetism
Replies
25
Views
1K
• Electromagnetism
Replies
15
Views
1K
• Electromagnetism
Replies
1
Views
656
• Electromagnetism
Replies
30
Views
2K
• Electromagnetism
Replies
21
Views
2K
• Electromagnetism
Replies
3
Views
1K
• Electromagnetism
Replies
3
Views
880