Electric field using Gauss' Law

In summary, the two spheres have a net charge of -2.35 μC and 4.35 μC, respectively. The charge is uniformly distributed within the volume of each sphere. When the spheres are placed halfway between each other, the electric field between the two spheres is +0.14 V/m.
  • #1
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Homework Statement


Two small insulating spheres with radius 9.00×10−2m are separated by a large center-to-center distance of 0.545 m . One sphere is negatively charged, with net charge -2.35 μC , and the other sphere is positively charged, with net charge 4.35 μC . The charge is uniformly distributed within the volume of each sphere.
What is the charge at the midway point between charged spheres

Homework Equations


Gauss' Law: I will use ∫ for closed surface integral
∫E⋅dA=Qenclosed/ε0

The Attempt at a Solution


I used two gaussian surfaces. One encapsulating the negatively charged sphere and one the positively charged sphere. The radius of my gaussian surface was half the distance between the spheres, 0.2725m.
In gauss' law E can come out of the integral and the integral of dA is A.
Area of sphere is 4πr2
E4π(0.2725m)2=(2.35x10-6C)/(8.85x10^-12C2/Nm2)
This E will be pointing away from the negatively charged sphere at the midpoint between two charged spheres.

I did this for the positively charged sphere as well using a gaussian surface with the same radius(as above). I obtained an Electric field that this sphere contributes at the midpoint. I summed them up and got a total electric field. This answer is incorrect.
Not sure what I am doing wrong,
The problem gives the radius of both charged spheres so maybe this comes into play somewhere?
Thanks a bunch!
 
Last edited:
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  • #2
How do you imagine that we will be able to help you if you do not show us what you did? You need to both describe it in words and show us all your computations. Otherwise it is just not feasible to find where you go wrong.

Edit: You also failed to actually specify what the question asks you to do. You just described the setup.
 
  • #3
Orodruin said:
How do you imagine that we will be able to help you if you do not show us what you did? You need to both describe it in words and show us all your computations. Otherwise it is just not feasible to find where you go wrong.

Edit: You also failed to actually specify what the question asks you to do. You just described the setup.
I didnt write out all my calculations because I was more asking if the method was correct. What I mean is with a problem like this would drawing two gaussian surfaces, one around each charge, be a good place to start?
 
  • #4
Orodruin said:
How do you imagine that we will be able to help you if you do not show us what you did? You need to both describe it in words and show us all your computations. Otherwise it is just not feasible to find where you go wrong.

Edit: You also failed to actually specify what the question asks you to do. You just described the setup.
Just fixed question, my bad.
 

1. What is Gauss' Law?

Gauss' Law is a fundamental law in electromagnetism that relates the electric field at a point to the charge enclosed by a Gaussian surface surrounding that point.

2. How is Gauss' Law used to calculate electric fields?

Gauss' Law is used to calculate electric fields by simplifying the calculation of the electric field from a complex distribution of charges to a simple integral over the Gaussian surface.

3. What is a Gaussian surface?

A Gaussian surface is an imaginary surface that is used to enclose a charge or group of charges in order to apply Gauss' Law.

4. Can Gauss' Law be used for any charge distribution?

Yes, Gauss' Law can be used for any charge distribution, as long as the electric field is radially symmetric about the charge or group of charges enclosed by the Gaussian surface.

5. What are the units of electric field calculated using Gauss' Law?

The units of electric field calculated using Gauss' Law are newtons per coulomb (N/C) or volts per meter (V/m).

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