# Gauss: Conducting Spherical Shell w/ Point Charge

• calvert11
In summary, the problem involves a conducting spherical shell with inner and outer radii of 0.8 m and 1.3 m, respectively, and a net charge of 3 μC. There is also a point charge of 3 μC at the center of the hollow cavity. The task is to find the flux through a spherical Gaussian surface of radius 0.4 m in units of N · m2/C. Using Gauss's law and the formula for electric field, the solution is found to be 3.389e5 N*m2/C.
calvert11

## Homework Statement

Consider a conducting spherical shell with inner radius 0.8 m and outer radius 1.3 m. There is a net charge 3 μC on the shell. At its center, within the hollow cavity, there is a point charge 3 μC.

Determine the flux through the spherical Gaussian surface S, which has a radius of 0.4 m. Answer in units of N · m2/C.

a= 0.8m
b= 1.3m
q1 = q2 = 3e-6 C

Flux = E*A
E = (kQ)/r^2
A = 4pir^2

## The Attempt at a Solution

I thought I could approach this in a very straight forward way: finding E of the point charge and multiplying it by the spherical shell's surface area (using the outer radius for both r's).

So, E = (8.99e9*3e-6)/1.3^2 = 15958.6
A = 4*pi*1.3^2 = 21.237

EA = 3.389e5 N*m2/C, which is wrong. I'm not sure what I should do.

Your solution looks good to me. (Not sure why you used the outer radius, but it doesn't matter. For that matter, you could have just used Gauss's law directly.)

Your approach of finding the electric field of the point charge and multiplying it by the surface area of the spherical shell is a good start. However, there are a few things that need to be taken into consideration in order to get the correct answer.

Firstly, since the point charge is located at the center of the spherical shell, the electric field at any point within the hollow cavity will be zero. This is because the electric field inside a conductor is always zero due to the redistribution of charges on the surface. Therefore, the flux through the spherical Gaussian surface will also be zero.

Secondly, in order to find the electric field at the surface of the spherical shell, we need to take into account the electric field due to both the point charge and the net charge on the shell. This can be done by using the superposition principle, which states that the total electric field at any point is the vector sum of the electric fields due to each individual charge.

Therefore, the correct approach would be to find the electric field at the surface of the spherical shell due to the point charge and the net charge separately, and then add them together to get the total electric field at the surface. This total electric field can then be multiplied by the surface area of the spherical shell to get the flux through the Gaussian surface.

Using the equations given in the homework, the correct approach would be:

E1 = (k*q1)/r1^2 = (8.99e9*3e-6)/(0.4)^2 = 168712.5 N/C

E2 = (k*q2)/r2^2 = (8.99e9*3e-6)/(1.3)^2 = 15958.6 N/C

E_total = E1 + E2 = 184671.1 N/C

Flux = E_total * A = 184671.1 * 4*pi*(1.3)^2 = 3.813e6 N*m2/C

Therefore, the flux through the spherical Gaussian surface is 3.813e6 N*m2/C.

## What is Gauss's Law?

Gauss's Law is a fundamental law in electromagnetism that relates the electric field to the charge distribution in a given space. It states that the electric flux through a closed surface is proportional to the total charge enclosed by that surface.

## What is a spherical shell?

A spherical shell is a surface or object that is shaped like a hollow sphere. In the context of Gauss's Law, it refers to a hypothetical spherical surface that surrounds a point charge.

## How do you apply Gauss's Law to a spherical shell with a point charge?

To apply Gauss's Law to a spherical shell with a point charge, you need to calculate the electric field at any point outside the shell by using the formula E = kQ/r^2, where k is the Coulomb's constant, Q is the charge of the point charge, and r is the distance from the point charge to the point where you want to calculate the electric field.

## What is the significance of using a spherical shell in Gauss's Law?

Using a spherical shell in Gauss's Law allows for simplification of the calculations and makes it easier to determine the electric field at points outside the shell. This is because the electric field at any point outside the shell depends only on the total charge enclosed by the shell, not on the distribution of the charge.

## Can Gauss's Law be applied to non-spherical shells?

Yes, Gauss's Law can be applied to any closed surface, including non-spherical shells. However, the calculations become more complicated and may require the use of vector calculus. Spherical shells are often used in introductory examples because they simplify the calculations.

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