Electrostatic Potential

  • Thread starter Alex_Neof
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  • #1
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Homework Statement



What is the electrostatic potential for the electric field in the region x ≥ 0 where:

E
= (E0)*e^(kxi) V/m, with E0 a constant?

(The potential at x → infinity is defined to be zero).

Homework Equations


b
v = v_a - v_b = - ∫ E . dl
a


The Attempt at a Solution



0
v = - ∫ (E0)*e^(kx) dx = (E0)/k.
infinity

Is this correct?

Kind regards.
 

Answers and Replies

  • #2
BvU
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Helo Alex,

You can check for yourself if this is correct: how do you derive an electric field from a potential ? Does that work OK for your answer ?

Oh, and did I miss a minus sign somewhere ?
 
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  • #3
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Thank you BvU !
 
  • #4
BvU
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Does that mean you found a potential ##E_o\over k## V gives an electric field ## \vec E = E_0 \; e^{kx} \; {\bf \hat\imath} \ ## V/m ?
 
  • #5
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Does that mean you found a potential ##E_o\over k## V gives an electric field ## \vec E = E_0 \; e^{kx} \; {\bf \hat\imath} \ ## V/m ?
 
  • #6
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So I was given the electric field:

##\overrightarrow{E} = {E_0} e^{-kx}\hat i. ##

Which is completely in the x-direction.

Now using:

## V = {V_a} - {V_b} = - \int_a^b \overrightarrow{E}.\overrightarrow{dl}, ##

with the above electric field we have:

## V = - \int_\infty^x {E_0} e^{-kx}dx = \frac{E_0} {k} e^{-kx}.##

Then using what BvU suggested:
Helo Alex,

You can check for yourself if this is correct: how do you derive an electric field from a potential ? Does that work OK for your answer ?

Oh, and did I miss a minus sign somewhere ?
I remembered from my notes that the electric field is related to the potential as follows:

##\overrightarrow{E} = - \nabla V.##

So to test if the potential I found is correct, I just plug it in and find:

##- (\frac{\partial V} {\partial x})=-(-{E_0} e^{-kx}\hat i) ##

Which gives us back our original Electric Field:

##\overrightarrow{E} = {E_0} e^{-kx}\hat i.##

Thanks again BvU, much appreciated!
 
  • #7
BvU
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well done !
 
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  • #8
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Initially my limits for integration were wrong, I had them from ##\infty## to 0.
 

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