# Homework Help: Electric Potential and Equipotential Contour

1. Feb 10, 2012

### btpolk

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

The electric potential V in a region of space is given by the following expression, where A is a constant.
V(x, y, z) = A (6x^2 - 7y^4 + 6z^2)

(a) Derive an expression for the electric field vector E at any point in this region. (Use the following as necessary: A, x, y, and z.)

x-component
Ex = -12Ax

y-component
Ey = 28Ay^3

z-component
Ez = -12Az

(b) The work done by the field when a 2.50 µC test charge moves from the point (x, y, z) = (0, 0, 0.380 m) to the origin is measured to be 4.00 10-5 J. Determine A.

A = 18.47

(c) Determine the electric field at the point (0, 0, 0.380).

E(0, 0, 0.380) = 82.4 N/C negative z

(d) What is the radius of the equipotential contour corresponding to V = 1150 V and y = 9.00 m?

r =?????

2. Relevant equations

3. The attempt at a solution

All I need is part (d). I have no idea what it is talking about.

Last edited: Feb 10, 2012
2. Feb 10, 2012

### Sefrez

Being that electric potential can be related by ΔV = - ∫E ds = - ∫E_s ds, where E_s is the component of E in the direction of ds, you find that E_s = -dΔV/ds. Or simply, you can find the magnitude of the electric field in the direction of ds by taking the negative derivative of the electric potential. Likewise, to find x, y, and z components, you can take partial derivatives for each component. E = -dΔV/dx i -dΔV/dy j -dΔV/dz k

3. Feb 10, 2012

### btpolk

How do I use that to find the radius?

4. Feb 10, 2012

### Sefrez

Oh, sorry, I guess I misread your post. I assume what you are doing is finding the radius at y = 9 m made on the plane by x and z. Or that is, a "circle" equipotential field exists for all y on the x,z plane in this particular electric potential function.

So:
1150 = A (6x^2 - 7(9)^4 + 6z^2)
1150/A = 6x^2 + 6z^2 - 45927
45927 + 1150/A = 6x^2 + 6z^2
(45927 + 1150/A)/6 = x^2 + z^2
(45927 + 1150/18.47)/6 = x^2 + z^2

Where this is remembered to be a elliptical equation where both the major and minor axis's are equal (hence a circle, not an ellipse.) And thus:
r^2 = (45927 + 1150/18.47)/6
r = 87.55

Someone can double check if I did that correctly.

5. Feb 10, 2012

### btpolk

Yeah I just submitted the answer and it checks out. I guess I was just over thinking it and I didn't even know what a equipotential contour was...

6. Feb 10, 2012

### Sefrez

Yes, an equipotential surface is a surface where the electric potential is the same. Any point on the surface has the same potential difference.

7. Feb 10, 2012

### btpolk

I appreciate the help! I do have another question posted if you have some free time