Coulomb's Law and Gauss's law

  • Thread starter henrybrent
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Two electric charges q1 = 2 microC and q2 = -2 microC are located at [itex]\vec{r_1}=(1,0,0)m[/itex] and [itex]\vec{r_2}=(0,0,-1)m[/itex] respectively. Calculate the force on q1 in vector form.


[itex] \vec{F_{12}} = \frac {1}{4\pi\epsilon_0}\frac{q_1q_2}{|\vec{r_1}-\vec{r_2}|^3}(\vec{r_1}-\vec{r_2}) [/itex]

is the formula I am using.

I get an answer of (-0.013, 0, -0.013)N but I don't think this is correct and I have no idea where I messed up.

any ideas? ( I have no numerical solutions to the questions)

Calculate the electric dipole moment of this system (vector form)

I have used P = qD

q = 2x10^-6
d = (1,0,1) which I obtained from using r1-r2

So I just get the charge again multiplied by the vector

Don't think this is correct either.

Calculate the electric fluxes through two spherical Gaussian surfaces centred at the origin with radii R1=0.5m and R2=10m respectively

Now this is confusing. The first sphere doesn't enclose any charge, so q_enclosed is just 0? And the 10m sphere encloses both, but when you find the total charge enclosed, it's 2+ (-2) = 0? so again 0?

Any help appreciated
 

Answers and Replies

  • #2
haruspex
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All looks good to me. What makes you think it's wrong?
 
  • #3
nasu
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How can someone tell you where you messed up if you don't show your work?
 
  • #4
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[itex] \vec{F_{12}} = \frac {1}{4\pi\epsilon_0}\frac{(2x10^-6)(-2x10^-6)}{\sqrt{2}^3}(1,0,1) [/itex]

That is my working
 
  • #5
nasu
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It looks OK. So does your result.
 

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