Gauss's Law and charge distribution

In summary: But I think you have a typo in b) where you have 10^1 for your result. It should be 10^-9 if you use the same units as in c).
  • #1
KillerZ
116
0

Homework Statement



A spherically symmetric charge distribution produces the electric field [tex]E = (5000r^2) N/C[/tex], where r is in m.

a) What is the electric field strength at r = 20cm?

b) What is the electric flux through a 40cm diameter spherical surface that is concentric with the charge distribution?

c) How much charge is inside this 40cm diameter spherical surface?

Homework Equations



[tex]E = (5000r^2) N/C[/tex]

The Attempt at a Solution



I am not sure if I did this right?

a)

[tex]E = (5000r^2) N/C[/tex]

[tex]E = (5000(0.2m)^2) N/C[/tex]

= [tex]200 Nm^2/C[/tex]

b)

40cm = diameter
20cm = radius

[tex]E = (5000(0.2m)^2) N/C[/tex]

= [tex]200 Nm^2/C[/tex]

c)

[tex]\phi_{e}=\frac{q}{\epsilon_0}[/tex]

[tex]q=(\phi_{e})(\epsilon_0)[/tex]

[tex]q=(200 Nm^2/C)(8.85x10^{-12} C^2/Nm^2)[/tex]

[tex]= 1x10^{-9} C[/tex]
 
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  • #2
You are stumbling on the units. They should have written E=5000*(r/m)^2*N/C, in my opinion. I think that's what they really mean by "where r is in m". Now if you do a) you get E=5000*(0.2m/m)^2*N/Q. Now you have correct E field units of N/C. For the flux in b) you need to integrate that field over the surface of the sphere. Now you pick up the extra m^2 in the units. But you have to multiply the E field from a) by the area of the sphere.
 
  • #3
Ok, I redid it:

a)

[tex]E = (5000r^2) N/C[/tex]

[tex]E = (5000(0.2m/m)^2) N/C[/tex]

= [tex]200 N/C[/tex]

b)

40cm = diameter
20cm = radius

[tex]\phi_{e}=\oint EdA = EA_{sphere}[/tex]

= [tex](200 N/C)(4\pi(0.2m)^2)[/tex]

=[tex]101 Nm^2/C[/tex]

c)

[tex]\phi_{e}=\frac{q}{\epsilon_0}[/tex]

[tex]q=(\phi_{e})(\epsilon_0)[/tex]

[tex]q=(101 Nm^2/C)(8.85x10^{-12} C^2/Nm^2)[/tex]

[tex]= 8.9x10^{-10} C[/tex]
 
  • #4
That looks much better.
 

Related to Gauss's Law and charge distribution

1. What is Gauss's Law and how is it related to charge distribution?

Gauss's Law is a fundamental law of electromagnetism that describes the relationship between electric charges and electric fields. It states that the electric flux through a closed surface is equal to the charge enclosed by that surface divided by the permittivity of free space. This law is directly related to charge distribution because it allows us to calculate the electric field at any point in space by knowing the distribution of charges.

2. How is charge distribution related to electric potential?

The distribution of charges in a system affects the electric potential at any point in space. In systems with a uniform charge distribution, the electric potential decreases linearly with distance from the source. In contrast, in systems with non-uniform charge distributions, the electric potential is more complex and can vary significantly depending on the distribution of charges.

3. What is a Gaussian surface and how is it used in Gauss's Law?

A Gaussian surface is a mathematical construct used to simplify the application of Gauss's Law. It is an imaginary closed surface that is chosen such that the electric field is constant and perpendicular to the surface at every point. This allows us to calculate the electric flux through the surface by simply multiplying the magnitude of the electric field by the area of the surface. This simplifies the calculation of electric fields for complex charge distributions.

4. What is the difference between a continuous and discrete charge distribution?

A continuous charge distribution refers to a system where the charges are spread out over a continuous region, such as a wire or a plate. In contrast, a discrete charge distribution refers to a system where the charges are localized at specific points, such as in a collection of point charges. The application of Gauss's Law differs for these two types of distributions, as the integration over a continuous distribution is replaced by a summation over discrete charges.

5. Can Gauss's Law be applied to systems with non-uniform charge distributions?

Yes, Gauss's Law can be applied to systems with non-uniform charge distributions. However, the calculation of the electric field in these systems can be more complex and may require advanced mathematical techniques. In some cases, it may be necessary to break the system into smaller subsections with uniform charge distributions in order to apply Gauss's Law and calculate the electric field at a specific point.

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