# little question on potential

• Mar 20th 2007, 08:18 PM
little question on potential
Hi,
I am a little bit confused with how to find the electric potential everywhere.

I have this electric field:
E=x(x^2)+y(y^2)+z(2*z) where x,y,z are unit vectors.

Find the electric potential everywhere?

I know that V=-[integral] E*dL

But the fact that it is said "everywhere" confused me. I understand better when everywhere is used with two concentric cylinders or spheres but in the plane like this ..??

I guess I just have to do the integration without boundary.
Please, can someone tell me what I can do?
• Mar 20th 2007, 10:47 PM
CaptainBlack
Quote:

Hi,
I am a little bit confused with how to find the electric potential everywhere.

I have this electric field:
E=x(x^2)+y(y^2)+z(2*z) where x,y,z are unit vectors.

Find the electric potential everywhere?

I know that V=-[integral] E*dL

But the fact that it is said "everywhere" confused me. I understand better when everywhere is used with two concentric cylinders or spheres but in the plane like this ..??

I guess I just have to do the integration without boundary.
Please, can someone tell me what I can do?

What is wrong with V(x,y,z) = (1/3)(x^3+y^3+z^3)?

RonL
• Mar 21st 2007, 07:10 AM
Nothing I guess. Thanks
• Mar 21st 2007, 10:53 AM
CaptainBlack
Quote:

Originally Posted by CaptainBlack
What is wrong with V(x,y,z) = (1/3)(x^3+y^3+z^3)?

RonL

Opps, there is an error in the last term, I read the z component of the field
as being z^2, when in fact it is 2*z, so we should have:

V(x,y,z) = (1/3) x^3 + (1/3) y^3 + (1/4) z^2

RonL
• Mar 21st 2007, 02:08 PM
ecMathGeek
Quote:

Originally Posted by CaptainBlack
Opps, there is an error in the last term, I read the z component of the field
as being z^2, when in fact it is 2*z, so we should have:

V(x,y,z) = (1/3) x^3 + (1/3) y^3 + (1/4) z^2

RonL

lol... sorry to do this to you, CaptainBlack, but as you said, it's a 2*z, therefore, INT(2z)dz = 2(1/2)z^2 = z^2

V(x,y,z) = (1/3) x^3 + (1/3) y^3 + z^2

But since braddy said that V=-[integral] E*dL, it's the negative integral, so:

V(x,y,z) = -(1/3) x^3 - (1/3) y^3 - z^2
• Mar 21st 2007, 05:54 PM
topsquark
Quote:

Hi,
I am a little bit confused with how to find the electric potential everywhere.

I have this electric field:
E=x(x^2)+y(y^2)+z(2*z) where x,y,z are unit vectors.

Find the electric potential everywhere?

I know that V=-[integral] E*dL

But the fact that it is said "everywhere" confused me. I understand better when everywhere is used with two concentric cylinders or spheres but in the plane like this ..??

I guess I just have to do the integration without boundary.
Please, can someone tell me what I can do?

The previous answers are correct. What you need to realize is that, since the electric potential is a "conservative" function, the value of the integral does not depend on the path taken, so you may choose any path you wish. The "simplest" path to calculate in this case would be one where you go from the origin (I'm starting here since V(0, 0, 0) = 0 V) straight to (x, 0, 0) straight to (x, y, 0) straight to (x, y, z). You will obtain ecMathGeek's answer.

-Dan