Electric Potential

Electric Potential due to

Continuous Charge Distributions When we looked at electric fields, we developed Coulomb's Law for the field from a point charge and then used that to find the electric field due to a continuous distribution of charge. Now, as we look at electric potential, we do the same thing. However, this time it will be easier since electric potential is ascalar. No directions or components are required.Consider a point P in space. What is the electric potential there because of a continuous distribution of charge? For some little amount of charge dq, there is a contribution to the electric potential at P; that contribution dV is

dV = k (dq / r)

where r is the distance from point P to our little amount of charge dq. Typically dq will be a charge density multipled by a differential length, area, or volume. That is, we will usually have

dq = ds

dq = dA

or

dq = dV

Be careful,

thisdV is a small, differential volume like dV = dx dy dz. We arealsousing V forelectric potentialso dV usually means the contribution to the electric potential due to some small differential charge dq. Be careful!We can write

But, of course, the "fun" or the work or the understanding comes from the details that depend upon the geometry of a particular situation. Therefore, let us look at some particular examples:

Consider a

ringof charge. The ring has radius a and we want to calculate the electric potential at a point P on the axis of the ring, a distance x away from the ring.

Now that we can handle aringof charge, let us extend that geometry and integrate the ring of charge from r = 0 to r = a and use adiskof charge.For a ring of radius r and thickness dr, the charge dq on that ring is the charge density sigma multiplied by the area

dq = [ dA ] dq = [ 2 r dr ]

Make a variable substitution, u = x

^{2}+ r^{2}so du = 2 r dr,Don't forget the

limits!

Consider a

line of charge:Now the small differential charge dq is the charge density multiplied by a differential length dx

dq = dx If we have a line of charge with a length L, we will need to integrate dx from x = 0 to x = L.

Consider a line of charge as sketched here. What is the electric potential at a point P a distance d away from the end of the line of charge of length L, carrying total charge Q uniformly distributed along its length?

For a small differential charge dq located at position x, the distance to point P is r where

For such a point charge dq located a distance r away from P, we know it contributes dV to the potential at P,

We can look this up in a table of integrals and find

But

don't forgetthelimits!

We have calculated the electric potential. We could use these expressions andE = - dV/dx or

E = - dV/dr to calculate the electric field from these distributions.

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