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Gauss's Law and charge distribution

  • Thread starter KillerZ
  • Start date
  • #1
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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]
 

Answers and Replies

  • #2
Dick
Science Advisor
Homework Helper
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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
116
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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
Dick
Science Advisor
Homework Helper
26,258
618
That looks much better.
 

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