Spherical surface surrounding charges find flux

In summary, the electric flux through a spherical surface surrounding a collection of charges can be found by dividing the charge enclosed by the permittivity constant. The signs of the charges determine the direction of the flux. In the given example, the electric flux for a single +5.5 x 10-6 C charge is 6.117 x 10^-16 and for a single -4.8 x 10-6 C charge it is -5.369 x 10^-16. For both charges, the total flux is 7.46 x 10^-17.
  • #1
phys62
53
0
A spherical surface completely surrounds a collection of charges. Find the electric flux (with its sign) through the surface if the collection consists of (a) a single +5.5 x 10-6 C charge, (b) a single -4.8 x 10-6 C charge, and (c) both of the charges in (a) and (b).

Ok, so this is what I've been able to figure out:
Ea=kq1/r^2=49445/r^2

Eb=kq2/r^2=-43152/r^2

and for part c: Ea + Eb = 6293/r^2

I'm just so confused as to how to get rid of the r^2.. Have I used the wrong equation entirely? Thanks so much to anyone who can help!
 
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  • #3
The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.

so then for example on part a, would I not just take:

q/k = (5.5x10^-6)/(8.99x10^9) = 6.117x10^-16

?? I know that's not right but I don't understand why. Do I have the sign backwards?
 
  • #4
phys62 said:
The total of the electric flux out of a closed surface is equal to the charge enclosed divided by the permittivity.

so then for example on part a, would I not just take:

q/k = (5.5x10^-6)/(8.99x10^9) = 6.117x10^-16

?? I know that's not right but I don't understand why. Do I have the sign backwards?

Φ = Q/εo

But you used the right value.

If the charge is - then the flux is -, meaning that it points towards the charge. If + it would be pointing outward through the closed surface.
 
  • #5
so you're saying my answer is right then, right? I submitted it on my online homework thing, and it says I'm incorrect :/
 
  • #6
phys62 said:
so you're saying my answer is right then, right? I submitted it on my online homework thing, and it says I'm incorrect :/

Do they want 6.1 instead of 6.117, since they only gave you 5.5 for the charge in the first place?

I ask in case this is something your instructor likes to do.
 
  • #7
No, our answers can be to whatever decimal point we want, just as long as it's within +/- 2% of the correct answer
 
  • #9
Ohhh. lol. okay, now if only I knew what eo was..
 
  • #10
oh wait... ok so the equation is Φ = Q/εo

do I just need to solve for Eo?
 
  • #11
I see! lol thanks so much for your help! :]
 
  • #12

Related to Spherical surface surrounding charges find flux

What is a spherical surface?

A spherical surface is a three-dimensional shape that is perfectly round, like a ball or a globe. It is defined by a fixed radius and all points on the surface are equidistant from the center.

What are charges?

Charges are fundamental properties of matter that can be either positive or negative. They are responsible for the attractive or repulsive forces between objects and are expressed in units of coulombs (C).

What is flux?

Flux is a measure of the flow of a quantity through a surface. In the context of electromagnetism, it represents the amount of electric field passing through a given area.

How do you find the flux through a spherical surface surrounding charges?

To find the flux through a spherical surface surrounding charges, you can use the Gauss's Law, which states that the electric flux through a closed surface is proportional to the net electric charge enclosed by the surface. This means that the flux can be calculated by dividing the net charge enclosed by the surface by the permittivity of free space and multiplying by the surface area.

Why is finding the flux through a spherical surface surrounding charges important?

Finding the flux through a spherical surface surrounding charges is important because it allows us to understand and analyze the behavior of electric fields. It also helps in predicting the strength and direction of the electric field at various points in space, which is crucial in many practical applications such as designing electrical circuits and devices.

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