Magnitude of electric flux through a sphere?

In summary, the electric field on the surface of a 6.0 cm -diameter sphere is perpendicular to the surface of the sphere and has a magnitude of 52 kN/C.
  • #1
jlmccart03
175
9
****EDIT****: I had improper significant figures. It was the correct number.
1. Homework Statement

The electric field on the surface of a 6.0 cm -diameter sphere is perpendicular to the surface of the sphere and has magnitude 52 kN/C .

What is the magnitude of the electric flux through the sphere? Notice that the units of the answer involve kilo-Newtons.

Homework Equations


I = EA or I = E4pir2 = Q/epsilon

The Attempt at a Solution


I first attempted to try and use Gauss's Law, but didn't get as far as I had hoped since we are not given a charge as thought previously.

I then decided to us E*4pir2 to get 5881.06 kN *cm2/C which is 0.588106 kN*m2/C which is wrong. So I basically do not know how to approach this problem.

Thanks in advance for any help!
 
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  • #2
The problem statement seems to be very incomplete because it doesn't say where the electrical charge is that is creating this electric field. You will get a different answer depending on the distribution of this charge. e.g. is it a surface charge, is it uniformly distributed throughout the sphere, or is there a point source at the center of the sphere?
 
Last edited:
  • #3
Charles Link said:
The problem statement seems t be very incomplete because it doesn't say where the electrical charge is that is creating this electric field. You will get a different answer depending on the distribution of this charge. e.g. is it a surface charge, is it uniformly distributed throughout the sphere, or is there a point source at the center of the sphere?
My answer was correct, but was missing a few significant figures. I think in general it was just wanting us to use Gauss's Law on the most simplified version possible. Indeed it is incomplete, but that was everything the question gave me unfortunately.
 
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  • #4
Charles Link said:
The problem statement seems t be very incomplete because it doesn't say where the electrical charge is that is creating this electric field.
If the magnitude of the flux is uniform over the entire surface of the sphere then we can assume a spherically symmetric charge distribution that can be then be treated as a point charge located at the sphere's center.
jlmccart03 said:
My answer was correct, but was missing a few significant figures.
Strange, as I get quite different digits when I do the same calculation.
 
  • #5
gneill said:
If the magnitude of the flux is uniform over the entire surface of the sphere then we can assume a spherically symmetric charge distribution that can be then be treated as a point charge located at the sphere's center.

Strange, as I get quite different digits when I do the same calculation.
Huh? Really? This is a screenshot of the question and answer.

upload_2017-1-30_17-42-49.png
 
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  • #6
My apologies. I mistook the diameter value to be a radius :oops: I hate when that happens :mad:

Anyhow, it's great that you arrived at the correct result. Cheers!
 

Related to Magnitude of electric flux through a sphere?

What is the formula for calculating the magnitude of electric flux through a sphere?

The formula for calculating the magnitude of electric flux through a sphere is Φ = E * A * cos(θ), where Φ is the magnitude of electric flux, E is the electric field strength, A is the surface area of the sphere, and θ is the angle between the electric field and the surface normal.

What is the unit of measurement for the magnitude of electric flux?

The unit of measurement for the magnitude of electric flux is newtons per meter squared (N/m^2) or volts per meter (V/m). This unit can also be written as coulombs per meter squared (C/m^2) since electric flux is also equal to the electric field strength multiplied by the surface charge density.

How does the magnitude of electric flux through a sphere change with distance?

The magnitude of electric flux through a sphere decreases with distance from the center of the sphere. This is because the electric field strength decreases with distance according to the inverse square law. As the distance from the center of the sphere increases, the surface area of the sphere also increases, resulting in a smaller magnitude of electric flux.

Can the magnitude of electric flux through a sphere be negative?

Yes, the magnitude of electric flux through a sphere can be negative. This occurs when the electric field and the surface normal have an angle greater than 90 degrees, resulting in a negative cosine value in the formula for electric flux. A negative electric flux indicates that the electric field is directed into the surface of the sphere instead of outwards.

How can the magnitude of electric flux through a sphere be used in practical applications?

The magnitude of electric flux through a sphere can be used in various practical applications, such as calculating the electric field strength at a point on the surface of the sphere or determining the total charge enclosed by the sphere. It can also be used in the design and analysis of electrical systems, such as capacitors and antennas.

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