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Conductors in Electrostatic Equilibrium problem

  • Thread starter dimpledur
  • Start date
  • #1
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Homework Statement


In fair weather, over flat ground, there is a downward electric field of about 150 N/C. A) assume that the Earth is a conducting sphere with charge on its surface. If the electric field just outside is 150 N/C pointing radially inwards, calculate the total charge on the Earth's surface. B) at an altitude of of 250 m above the Earth's surface, the field is only 120 N/C. Calculate the charge density.


Homework Equations



[tex]E=\frac{k\left|q\right|} {r^2}[/tex]

[tex]\rho=\frac{q} {V}[/tex]

The Attempt at a Solution



I only need help with part b)

[tex]\left|q\right|=\frac{Er^2} {k}[/tex]

[tex]\left|q\right|=\frac{(120 N/C)[6.371(10)^6 m +250 m]^2} {8.99(10)^9 Nm/C^2}[/tex]

[tex]\left|q\right|=5.42(10)^5 C[/tex]

We then have,

[tex]\rho=\frac{5.42(10)^5 C} {(4/3)(\pi)[(6.371(10)^6 m +250 m)^3-(6.371(10)^6 m)^3]}[/tex]

[tex]\rho=4.25(10)^{-12} C/m^3[/tex]

However, the answer should be [tex]\rho=1(10)^{-12} C/m^3[/tex] according to the back of the book. Suggestions?
 

Answers and Replies

  • #2
1,137
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[tex]\rho[/tex] is surface charge density not volumetric.

Use [tex]\rho[/tex] = q/(surface area)
 
  • #3
194
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In my text book [tex]\rho[/tex] is for volumetric charge density.. Additionally, the answer is given per unit volume, so surface area is not suffice.
 
  • #4
1,137
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Calculate the charge density.
Who's density are they talking about?
 
  • #5
194
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I'm assuming that they want to know the charge density of the air
 
  • #6
1,137
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well if there's charge in air also and you need only density in air then net E at 120m is due to both charge on earth and air in between.

So find field due to earth at 250m above earth and subtract it form E given and then find ρ
 

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