# Electrostatic Potential

1. Apr 22, 2015

### Alex_Neof

1. The problem statement, all variables and given/known data

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).

2. Relevant equations
b
v = v_a - v_b = - ∫ E . dl
a

3. The attempt at a solution

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

Is this correct?

Kind regards.

2. Apr 22, 2015

### BvU

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 ?

3. Apr 24, 2015

### Alex_Neof

Thank you BvU !

4. Apr 24, 2015

### BvU

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. Apr 24, 2015

### BvU

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. Apr 25, 2015

### Alex_Neof

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:
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. Apr 25, 2015

### BvU

well done !

8. Apr 25, 2015

### Alex_Neof

Initially my limits for integration were wrong, I had them from $\infty$ to 0.