Evaluating Electric Field for the Coloumb Potential: A Guide

In summary, the problem is to prove that for an electric potential V(r) that depends only on the distance r=(x^2 + y^2 + z^2)^1/2 from the origin, the gradient is given by gradV(r) = V'(r)r-hat, where r-hat is a unit vector in the direction of r and V'(r) is the derivative of V(r) with respect to r. This can be used 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) and E0 stands for epsilon. To do this, we can take
  • #1
jlmac2001
75
0
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.
 
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  • #2
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.
 
  • #3
does this look right?

This is the answer i got:

kq*(x^2+y^2+z^2)^1/2 /(xi+yj+zk)
 
  • #4


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:
[tex]\frac{kq}{r^2}\hat{r}[/tex]
 
  • #5
i tried it again

this time i got, -kq(xi+yj+zk)/(r^5)
 
  • #6
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 [itex]\hat{r}[/itex] direction. So E = -grad(V)= -kq(∂/∂r)(1/r), which gives [itex]\frac{kq}{r^2}\hat{r}[/itex].

If you wish to use cartesian coordinates, no problem:
[tex]V = \frac{kq}{(x^2 + y^2 + z^2)^\frac{1}{2}}[/tex]
so, -grad(V) =
[tex]\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}[/tex]
 

1. What is the purpose of evaluating electric field for the Coulomb potential?

The purpose of evaluating electric field for the Coulomb potential is to calculate the strength and direction of the electric field at a given point in space due to a point charge. This is important in understanding how electric charges interact with each other and how they affect their surroundings.

2. How do you calculate the electric field for the Coulomb potential?

The electric field for the Coulomb potential is calculated using the formula E = kQ/r2, where E is the electric field strength, k is the Coulomb's constant, Q is the magnitude of the point charge, and r is the distance from the point charge to the point where the electric field is being calculated.

3. What units are used to measure the electric field for the Coulomb potential?

The electric field for the Coulomb potential is measured in units of Newtons per Coulomb (N/C) or Volts per meter (V/m). Both of these units represent the strength of the electric field at a given point in space.

4. Can the electric field for the Coulomb potential be negative?

Yes, the electric field for the Coulomb potential can be negative. This indicates that the electric field is directed in the opposite direction of the positive charge. A positive electric field indicates that the field is directed away from the positive charge.

5. How does the distance from the point charge affect the electric field for the Coulomb potential?

The electric field for the Coulomb potential is inversely proportional to the square of the distance from the point charge. This means that as the distance increases, the electric field strength decreases. In other words, the electric field is stronger at points closer to the point charge and weaker at points further away.

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