Electric Potential from Electric Field

[jaguar7]

Homework Statement

Electric Potential from Electric Field

Suppose that, as a function of x, y, and z, an electric field has the following components:
Ex = 6x^2y, Ey = 2x^3 + 2y and Ez = 0
where E is measured in V/m and the distances are measured in m. Find the electrical potential difference between the origin and the point x = -0.4 m, y = 6.1 m, z = 0.0 m.

Homework Equations

The Attempt at a Solution

We should be able to just do E dot r, which gives (6x^2y)x + (3x^3+2y)y, where x = -.5 and y = 6.1

this didn't work.

Reattempt:
for reference: Ex = 6x^2y, Ey = 2x^3 + 2y, x = -0.4 m, y = 6.1 m

dV = integral E dot dr from 0 to the point = integral Ex dx from 0 to -.4 + integral Ey dy from 0 to 6.1
= integral 6x^2y dx from x=0 to -.4 + integral 2x^3 + 2y from y=0 to 6.1

= y*2x^3 |(x=0 to -.4) + (2x^3y + y^2)|(y=0 to 6.1)

=2(-.4^3)y + (12.2x^3 + 6.1^2),

where x = -.4 and y=6.1

plug in and get 35.6V,

wrong...

 

[gneill]

When you're integrating along the path, from the origin to x = -0.4 the value of y should be zero (you're moving out along the x-axis first). After this, the value of x will be fixed at -0.41 as your path takes you vertically from y = 0 to y = 6.1 .

 

[jaguar7]

When you're integrating along the path, from the origin to x = -0.4 the value of y should be zero (you're moving out along the x-axis first). After this, the value of x will be fixed at -0.41 as your path takes you vertically from y = 0 to y = 6.1 .

How do you integrate it? It's a vector...

 

[gneill]

How do you integrate it? It's a vector...

There's a dot product involved. E and dr are both vectors, but their dot product is a scalar.

Now, you have a choice. You can do it the hard way, which is to take a straight-line path from the origin to the endpoint, developing an expression for E.dr for that trajectory, or you can realize that the electric field is conservative and that the work done going from point A to point B is independent of the path taken.

So to go from the origin to point (x,y), you can first take a straight line from the origin to x, keeping y = 0, and then from point (x,0) to (x,y) along the straight line where x is constant.

In the first leg of the journey dr is differential element that points in the x direction. In the second leg it points in the y direction. So dr is a vector <dr,0,0> in the first case, and <0,dr,0> in the second. This makes the dot products particularly simple, because dr simply "picks out" either the x or the y component of the field vector.