from Office of Academic Technologies on Vimeo.

**Example 2- Calculating electric field of a ring charge from its potential**

Let’s do an example for calculating the electric field from the potential, and let’s recall the ring charge. Earlier we calculated the ring charge potential, which was equal to *q* over 4*π* *ε**0* square root of *z*2 plus *R*2 for a ring with radius of big *R*, and the potential that it generates *z* distance away from its center along its axis and with a charge of positive *q* distributed uniformly along the circumference of the ring charge.

Now, knowing this potential, let’s try to figure out the electric field that it generates at this point. Again, if you recall, we calculated that electric field by applying Coulomb’s law earlier, and now we will follow a different approach. Assume that first we calculate the potential, which we did this also earlier through an example, and then from this potential we would like to figure out the electric field.

We have seen that the rate of change of potential with respect to distance gives the component of the electric field along that direction. In a rectangular coordinate system the *x* component of the electric field was the negative partial derivative of the potential with respect to *x* direction and with respect to *x*-coordinate and the *y* component was equal to -∂*v* over ∂*y* and the *z* component of the electric field was -∂*v* over ∂*z*.

Well, first if we try to calculate to *x* component of the electric field so we’ll take the partial derivative of this potential function with respect to *x*, ∂ over ∂*x* off *q* over 4*π* *ε**0* square root of *z*2 plus *R*2. This operation is going to give us 0 because there’s no *x* dependence in this expression and we keep all the other quantities as constant while we’re taking the derivative with respect to *x*. So that’s going to be equal to 0, due to the fact that no *x* dependence. And similarly ∂ over ∂*y*, the *y* component of the electric field which is minus ∂ over ∂*y* of the potential function *V* which will be also equal to 0 again due to the fact that in this case also there is no *y* dependence.

So the *x* component of the electric field is 0 for this case, *y* component is 0, and the only component left is the *z* component which is going to be equal to minus ∂ over ∂*z*, and now we can actually use the, if you want the total differentiation because there is no *x* and *y* dependence over here, *Q* over 4*π* *ε**0* square root of *z*2 plus *R*2.

Now to be able to take this derivative, first of all *Q* over 4*π* *ε**0* is constant we just take it outside of the derivative operator, and inside of the operator now we can express this with total differential, *d* over *dz* since there is no *x* and *y* dependence of lets move this square root in the denominator to the numerator and that’s going to make *z*2 plus *R*2 to the power of minus one-half. *z* component of the electric field will therefore be –*Q* over 4*π* *ε**0*.

If you take the derivative of this quantity, this function inside of the bracket with respect to *z,* we will have minus one-half times *z*2 plus *R*2. We will decrease the power by 1, so minus one-half minus 1 will give us minus 3 over 2 and now we will also take the derivative of the argument and since *R* is constant, derivative of *z*2 with respect to *z* is going to give us 2 *z*. Here this 2 and that 2 will cancel, minus and this minus will make positive, and the *z* component of the electric field will turn out to be *Q* times *z* — let’s move *z*2 plus *R*2 to the power minus 3 over 2 to the denominator — 4*π* *ε**0*, *z*2 plus *R*2 to the power of 3 over 2.

That tells us then the electric field is in *z* direction, or *d* because its *x* and *y* components are 0, and it has this magnitude. And if you go back and check it out with the example that earlier we did as we are calculating the electric field of a ring charge distribution along its axis by applying Coulomb’s law, you will see that we ended up with exactly the same result.