Before we begin our derivations for continuous charge distributions along charged rods, let's review some relationships developed in earlier lessons.

 

Recall that around every charged object there is an electric field, E, with field lines that originate on positive charges and terminate on negative charges. When another charged object is brought into a field it will experience a force of attraction or repulsion according to the formula.

 

F = qE

 

When we speak of the force between two point charges, we can calculate it magnitude with Coulomb's Law

 

 

The constant, k, in Coulomb's Law has the value

 

 

where

 

 

is called the permittivity of free space and represents the "willingness" of a region to establish an electric field. This value changes with the presence of dielectrics. We can now express Coulomb's Law as

 

 

Substituting this expression for F into our equation F = qE we get an equation that we can use to calculate the magnitude of the electric field around a point charge.

 

 

 

Suppose you need to calculate the electric field at point P located along the axis of a finite, uniformly charged rod. Let the charge distribution per unit length along the rod be represented by λ; that is,

 

 

The total charge represented by the entire length of the rod can consequently be expressed as

 

Q = λL.

 

 

 

 

The charge present on a small segment of the rod, Δx_i, can be expressed as Δq_i which equals 

Considering this small section to represent one of hundreds or thousands of point changes comprising the rod, its contribution to the electric field at P, , can be represented by

 

The total magnitude of the electric field at P would be equal to the sum of all these smaller contributions, Δ E_i.

 

Taking the limit as Δx approaches 0, we get that

 

 

where x = 0 is at point P. Integrating, we have our final result of

 

 

If the charge present on the rod is positive, the electric field at P would point away from the rod. If the rod is negatively charged, the electric field at P would point towards the rod. In either case, the electric field at P exists only along the x-axis.

 

Simplifying our expression for E_P further we note that as b becomes much greater than L, L + b approaches b and our formula for E_P returns to the more familiar expression for a point charge

 

where r = b.

 

 

 

 

 

 

Suppose you are now asked to calculate the electric field at point P located a distance b from the side of the uniformly charged rod.

 

 

Notice in the following diagram that we must deal with both the horizontal (E_x) and vertical (E_y) components of the electric field at P. Since the vertical components all point in the same direction -- away from the charged rod -- we will start with them.

 

 

Using vertical angles and right triangle trigonometry, we can calculate that

 

Taking the limit as Dx approaches 0, we get that

 

Unfortunately this leaves us with an expression involving three variables: x, r, and θ . Since our differentiable is dx, we need to replace r and θ with equivalent expressions involving only x. We can make this happen by noticing the following relationships:

 

Substituting for r and cos θ , we get

 

where x = 0 is at point O.

 

Since

After integrating, our final expression for E_y becomes

 

We will now repeat our process and solve for E_x.

 

Using vertical angles and right triangle trigonometry, we can calculate that

 

Again, this leaves us with an expression involving three variables: x, r, and θ . Since our differentiable is dx, we need to replace r and θ with equivalent expressions involving only x. We can make this happen by noticing the following relationships:

 

Substituting for r and sin θ , we get

 

Since

 

After integrating, our final expression for E_x becomes

 

Some interesting consequences