Example 1- Calculating electrical field of a disc charge from its potential
As another example to the obtaining the electric field from the potential, let’s recall the discharge potential. As you recall, when we consider a disc charge distribution with radius R and charge Q, with a positive charge Q, distributed uniformly throughout the surface, we calculated the potential that it generates along its axis, z distance away from its center at this point P, was equal to recall the potential, that this potential was equal to q over 2 π R2 ε0. Or let me just write it down as 2 π ε0 R2 times square root of z2 plus R2 minus z. This was the potential that it generated, and as you recall we obtained that by integrating the incremental potentials that the incremental concentric ring charges generated throughout the surface of this disc, and we obtained this expression.
Now, let us try to calculate the corresponding electric field at this point that the charge distribution generates from this potential. So, E at P is the question mark. Again, we know that the negative rate of change of potential with respect to distance gives us the component of the electric field along that direction. So in rectangular coordinate system, this was the x component of the electric field was equal to ∂V over ∂x. And in our case this is going to be equal to -∂ over ∂x of the potential function, which is in explicit form Q over 2 π ε0 R2 times square root of z2 plus R2 minus z.
Since there is no x dependence over here, this derivative is going to give us just 0. No x dependence. Similarly, Ey, which is going to be equal to -∂V over ∂y, will also give us 0 due to the fact that we have no y dependence in this equation too. And the only component left is the z component, which will be equal to minus ∂V over ∂z. And that will be equal to, since again, there is no x and y dependence, we can replace the partial derivative with the total derivative of d over dx of the potential function. And that is Q over 2 π ε0 R2 times square root of z2 plus R2 minus z, close parentheses.
z component, therefore, will be equal to Q over 2 π ε0 R2 is a constant, take it outside of the derivative operator, so we’ll have Q over 2 π ε0 R2. Now, let’s just go ahead and take the derivative. This is z2 plus R2 to the power one-half. If you take the derivative of that we will have one-half times z2 plus R2 and we will decrease the power by one. So we have plus one-half minus one, will give us minus one-half. And times the derivative of the argument, R is constant. Derivative of z2 will give us just 2z. That’s from the first term, and the second term will give us just 1. Derivative of z with respect to z is just 1. We’re taking the derivative with this factor z-coordinate and close parentheses.
Moving on, Ez will be equal to minus here we can cancel this 2 with this 2, and if we rewrite this expression, it’ll be Q over 2 π ε0 R2. Here z, and let’s move this z2 plus R2 to the denominator, therefore we will have z over square root of z2 plus R2 and minus 1. Let’s take one more step and put this negative into the bracket. In other words, multiply this this -1. Then Ez is going to be equal to Q over 2 π ε0 R2 times 1 minus z over square root of z2 plus R2.
Since x and y components of the electric field is 0, and we only have z component, that indicates that in that electric field is along the z-axis and it has this magnitude. Again, we did this example earlier. We calculated the electric field directly by applying Coulomb’s law, and if you compare the results of this example and that one, you will see that we have exactly the same expression.