Evaluating Electric Field for the Coloumb Potential

[jlmac2001]

The problem is:

Prove that the gradient of an electric potential V(r) which depends only on the distance r=(x^2 + y^2 + z^2)^1/2 from the origin has the vlaue gradV(r) = V'(r)r-hat where r-hat := r/r is a unit vector in the direction of r, and V'(r) := dV(r)/dr. Use this to evaluate the electric field E= -gradV(r) for the Coulomb potential
V(r)= kq/r from a point charge +q, where k=1/(4*pi*E0)

E0 stands for epsilon.

My question is:

I proved that gradV(r) = V'(r)r-hat. How do I evaluate the electric field for the Coloumb potential? Would I take the gradient of V(r)=kq/r? I don't get what I need to do.

 

[Doc Al]

Originally posted by jlmac2001
How do I evaluate the electric field for the Coloumb potential? Would I take the gradient of V(r)=kq/r?

Yes.

 

[jlmac2001]

does this look right?

This is the answer i got:

kq*(x^2+y^2+z^2)^1/2 /(xi+yj+zk)

 

[Doc Al]

Originally posted by jlmac2001
This is the answer i got:

kq*(x^2+y^2+z^2)^1/2 /(xi+yj+zk)

Check your work. Your answer must be equivalent to:
$$\frac{kq}{r^2}\hat{r}$$

 

[jlmac2001]

i tried it again

this time i got, -kq(xi+yj+zk)/(r^5)

 

[Doc Al]

I'm not sure what you're doing, but keep trying. You're probably making some simple error. Here's how I did it:

If you use spherical coordinates, the gradient is trivial to calculate. By symmetry, the gradient must be along the $\hat{r}$ direction. So E = -grad(V)= -kq(∂/∂r)(1/r), which gives $\frac{kq}{r^2}\hat{r}$.

If you wish to use cartesian coordinates, no problem:
$$V = \frac{kq}{(x^2 + y^2 + z^2)^\frac{1}{2}}$$
so, -grad(V) =
$$\frac{kq}{(x^2 + y^2 + z^2)^\frac{3}{2}}(x\hat{i}+y\hat{j}+z\hat{k}) = \frac{kq}{r^2}\hat{r}$$