# Gauss Law and Gauss surface [closed]

If we have hollow spherical conducting shell having no net charge. By placing positive point charge in the center of hollow conductor, negative charge will appear on inner face of conductor and positive charge on its outer surface, But what is its relation with Gauss law. This is common sense that negative charge will appear on its inner surface. What will be the electric field and electric flux inside and outside the shell?

## closed as off-topic by Jon Custer, John Rennie, sammy gerbil, heather, honeste_vivereAug 24 '17 at 14:47

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• What do you mean by applying a net positive charge to gauss surface and explain it in the form of Gauss law? – Kawin M Aug 20 '17 at 19:07
• I am doing struggle to understand Gauss law application. My textbook say that if we have hallow conductor having no net charge. By placing point positive charge in the center of hollow conductor, negative charge will appear on inner face of conductor and positive charge on its outer surface, But what is its relation with Gauss law. – Muhammad Numan Aug 20 '17 at 19:12
• Do you want to know the electric flux outside the hollow spherical conductor using gauss law? – Kawin M Aug 20 '17 at 19:22
• @MuhammadNuman, draw a spherical Gaussian surface inside the conductor, enclosing just the inner surface. Then, the flux through that surface is zero, because the electric field inside a conductor is zero. Apply Gauss law to find the charge density on the inner surface. – leongz Aug 20 '17 at 19:23
• Ok. And if i place a net positive charge on this hollow spherical ring. Where this net positive charge will be deposit? On inner surface or on its outer surface and what will be the reason of this? – Muhammad Numan Aug 20 '17 at 19:26

If positive charge is placed inside the hollow spherical shell

You actually don't even need gauss law to locate the charge on the outer surface of the hollow spherical shell.

Lets take a wire with two ends A and B. What happens if electrons from A moved to B, the end B will get negative charge and due to lack of electrons end A will get positive charge.

The same is with here, as the electrons moved towards the inner surface, the outer surface will get positive charge.

If no charge is placed inside the conductor and some electrons are removed from one of the surface

Ok now have you removed the net positive charge inside the hollow shell, so the conductor will be neutral now. Now, you remove some electrons from say outer surface of the shell, so this creates a positive charge in the outer surface. Soon after this happened, some electrons from inner surface will move to the outer surface and will make the conductor neutral again. Same vice versa.

To find electric field and electric flux on the surface of the shell

You are given a hollow spherical conducting shell with surface area A = 4(pi)R^2 (R is the radius of the shell) inside of which is a positive charge q. Now this q will induce a charge Q in the conductor. The magnitudes of these charges will be same q = Q. The surface charge density of shell will sigma = Q/A . The electric field at the surface of the shell will be E = sigma / E0. Electric flux will be Q/Eo.

To find electric field and electric flux outside the shell

Now if you want to know the electric field outside the shell at some point P at distance r from the charge q. Draw a spherical gaussian surface with radius r. Same as above, electric field at p in the gaussian surface will be E = Q / 4(pi) r^2 E0 . Electric flux will be Q/E0

• There are applications of Gauss law in my text book. One of them is related to spherical rings but i really don't have any idea what it say. I also watch Feynman lectures but still struggling to understand it. In simple words, What is the relation of Gauss with above diagram in question. – Muhammad Numan Aug 20 '17 at 19:45
• Ok now I got what you actually want. You are given a hollow spherical conducting shell with surface area A = 4(pi)R^2 (R is the radius of the shell) inside of which is a positive charge q. Now this q will induce a charge Q in the conductor. The magnitudes of these charges will be same q = Q. The surface charge density of shell will sigma = Q/A . The electric field at the surface of the shell will be E = sigma / E0. Electric flux will be Q/Eo – Kawin M Aug 20 '17 at 19:55
• Now if you want to know the electric field outside the shell at some point P at distance r from the charge q. Draw a spherical gaussian surface with radius r. Same as above, electric field at p in the gaussian surface will be E = Q / 4(pi) r^2 E0 . Electric flux will be Q/E0 – Kawin M Aug 20 '17 at 19:58

Gauss' law is useful for your problem because geometrically everything is spherically symmetrical. This in turn implies that the electric field will everywhere be spherically symmetrical.

Inside your conduction, the electric field is $0$. Thus, if you consider a point at location $\vec r$ inside your hollow sphere, and a Gaussian spherical of radius $r$, the flux of $\vec E$ through this Gaussian surface will be $0$ since $\vec E=0$ everywhere inside. Working backwards this tell you the net charge enclosed by your Gaussian sphere must be $0$. Since you have $+q$ at the center, there must be $-q$ elsewhere inside your Gaussian sphere. This $-q$ charge, by symmetry, must be uniformly distributed on the inner surface of your conductor. For future discussion, if $R_1$ is this inner radius, then the surface charge density on this inner surface will be $\sigma_1=-q/(4\pi R_1^2)$.

The field will change once you reach the outer radius. If your spherical conductor was initially neutral, and $-q$ is been distributed on the inner surface, it must be that $+q$ has (by induction), appeared on the outer surface of your conductor.

Suppose $R_2$ is the outer radius. Any Gaussian sphere of radius $r>R_2$ will then enclose a net charge of $+q$ since the total charge $-q$ on this inside surface is algebraically cancelled by the induced $+q$ charge on the outside, leaving a net charge of $+q-q+q=+q$ inside the Gaussian sphere. Therefore, by symmetry, the field will have spherical symmetry and $$\oint \vec E\cdot d\vec S = \vert E\vert 4\pi r^2 =\frac{q}{\epsilon_0} \tag{1}$$ from which you recover the field at $\vec r$: $$\vec E= \frac{q}{4\pi\epsilon_0 r^2}\hat r. \tag{2}$$ with $\hat r$ the unit radial vector in spherical coordinates.

Finally, note that even if a charge $+q$ has been induced on the outer surface, this charge is spread over a larger surface so the surface charge density $\sigma_2=q/(4\pi R_2^2)$ is actually smaller in magnitude than $\sigma_1$.

Gauss' law has allowed you to obtain $\sigma_1$, write (1) to deduce (2) and also indirectly obtain $\sigma_2$.