# Electric field at point Gauss law

• Pual Black
In summary, the conversation discusses the process of finding the electric field at point p due to a point charge q placed at o. The approach involves considering a sphere of radius r passing through point p and using Gauss law. However, there are mistakes in the given exercise and the correct method involves integrating 1/x3 to obtain the desired result.

## Homework Statement

We won't to find out the electric field at a point p due to a point charge q placed at o as shown in the figure
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Consider a sphere of radius r passing through the point p. Let the electric field at p be E then by Gauss law

my problem is how we get the last step? I don't understand how he integrate it.

## The Attempt at a Solution

i tried it and found a solution but not sure if its correct although i got the same result

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sorry that i don't used LATEX but I'm not used with it and i can't write the whole question with LATEX

You need to learn the rules for integration (and differentiation) of ##x^n##. Try this: Power Rule

Looks as if the power rule is known, but the concept of integration causes difficulties: substitution ##x=r^{-3}## is carried out ok initially (*), but the much better (or rather, the 'right') substitution ##x = r^{-2}##, leading to ##dx =-2\; r^{-3}dr## and hence ##\int r^{-3} dr = -{1\over2}\int dx## is overseen.

(*) Later on, the swap ##\int {x\over r^{-4}}\; dx = {1\over r^{-4}} \int x\; dx## is unforgivable: r is not a constant but depends on x !

If r is not constant and depends on x. Therefore i can't put the ##r^4## outside the integration. Right??

Indeed you can not !

Ok so how to get this solution?
Any idea??
I want the the same result ##E=\frac{q}{4 \pi \epsilon r^2}##

OK, we have to make a step back. Your "Consider a sphere of radius r passing through the point p. Let the electric field at p be E then by Gauss law " continues with a picture with formulas that seem to be out of context: at P there is no charge, so div E is zero there !

Where does this snippet come from? It also integrates from ## r=-\infty## to ## r=\infty## which is very weird, since r can't be negative. Then it equates that to ##2\int_0^\infty\ ##, then inserts a ##- {1\over 3 }## from nowhere (re-check the power rule)!

So, stepping back and trying to find the correct path:

Gauss law replaces a surface integral of E by a volume integral of rho. Indeed ##\nabla \vec E = {\rho\over \epsilon_0} ##, but ## \rho \ne q\big /{4\over 3}\pi r^3\ ## ! That is ##\rho## for a uniformly charged sphere with radius r !

In this simple case the volume integral of rho simply yields q. Frome there use rotation symmetry and gauss law to get to E.

Another thing: note that div E is not ## dE\over dr## !

Divergence in spherical coordinates tells us that

for rotational symmetry ##\nabla\cdot\vec E = {1\over r^2}\; {\partial (r^2E_r)\over \partial r}##

(
as you can check: div E = 0 for the desired result for E

and conversely: with rotational symmetry, Er2 is constant outside the area where charge is present.​
).

 forgot the divergence link, inserted it.

Pual Black said:
Ok so how to get this solution?
Any idea??
Sure. Learn how to integrate ##x^n##; in particular, ##\frac{1}{x^3} = x^{-3}##. (No fancy substitutions needed.)

Doc Al said:
Sure. Learn how to integrate ##x^n##; in particular, ##\frac{1}{x^3} = x^{-3}##. (No fancy substitutions needed.)

I know how to use the power rule.
If you see my solution ;) @BvU
Thank you for your help. Looks like this question is a mess. The whole question is wrong
I will now use your information and maybe i will get a right solution.

Pual Black said:
I know how to use the power rule.
If you see my solution
Then why did you say the following?
Pual Black said:
my problem is how we get the last step? I don't understand how he integrate it.
The last step was integrating 1/x3.

I guess I missed something. (Sorry about that!)

Yes that's right but if you try to integrate the last step you will not get the same result. Therefore i said that i don't understand it. But as BvU told me the whole question has mistakes.

So is the picture with the formulas from a different exercise ? Or did you do a part of the solution in TeX and then switch to handwritten ?

No the two pictures ( figure and Tex ) are from exercise. Just the handwriting is mine. But the exercise it is not from one source cause the exercise has the soltuion from a screenshot. It has text and image ( the soultion and the figure ) are screenshots.

So, do we have the "all clear"now, or is there still some thing unfinished ?

Thank you very much. Its clear now

## What is the electric field at a point according to Gauss's law?

The electric field at a point is defined as the force per unit charge experienced by a positive test charge placed at that point. According to Gauss's law, the electric field at a point is equal to the charge enclosed by a closed surface divided by the permittivity of free space.

## How is Gauss's law used to calculate the electric field at a point?

Gauss's law is used to calculate the electric field at a point by considering a closed surface surrounding the point and the charge enclosed within that surface. The electric field at the point is then determined by dividing the total charge enclosed by the permittivity of free space.

## What is the significance of Gauss's law?

Gauss's law is a fundamental law in electromagnetism that relates the electric field to the distribution of electric charges. It allows for the calculation of the electric field at a given point without having to consider the individual contributions of each charge. It also helps in understanding the principle of charge conservation.

## What are the assumptions made in using Gauss's law?

The assumptions made in using Gauss's law include having a closed surface surrounding the point of interest, the electric field being constant over the surface, and the charge distribution being symmetric. Additionally, the charges must be stationary and there should be no magnetic fields present.

## What are some practical applications of Gauss's law?

Gauss's law has a wide range of practical applications, including in the design of capacitors, electric motors, and other electronic devices. It is also used in understanding the behavior of lightning, electric fields in insulators, and the behavior of charged particles in accelerators.