(A) Suppose you need to calculate the electric field at point P located along the axis of a uniformly charged rod. Let the charge distribution per unit length along the rod be represented by l; that is, . The net charge represented by the entire length of the rod could then be expressed as Q = lL.

Let's first combine F = qE and Coulomb's Law to derive an expression for E.

For a small section Dx_i along the rod, the charge present will be Dq_i where

Dq_i = lDx_i

The electric field contribution at P by this section would be represented by

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

Taking the limit as Dx approaches 0, we get that

where x = 0 is at point P.

Integrating, we have our final result of

or

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.

(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 q. Since our differentiable is dx, we need to replace r and q with equivalent expressions involving only x. We can make this happen by noticing the following relationships:

Substituting for r and cos q, 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 q. Since our differentiable is dx, we need to replace r and q with equivalent expressions involving only x. We can make this happen by noticing the following relationships:

Substituting for r and sin q, we get

Since

After integrating, our final expression for E_x becomes

(D) Some interesting results to consider:

1. If we place point O at , then E_x = 0 since the left since the DE_x vectors cancel in corresponding pairs. This would leave us with only net E = E_y.

2. Suppose we view the bar from a very far distance; such that, and . Our expressions for E_x and E_y would reduce to