 Continuous Charge Distributions: Charged Rods and Rings Printer Friendly Version
 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. In all of the following cases, the line of charge is positive.
 Case I: P along the axis of a finite line of 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, Δxi, can be expressed as Δqi 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 Ei. 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 EP further we note that as b becomes much greater than L, L + b approaches b and our formula for EP returns to the more familiar expression for a point charge where r = b. Case II: P along the central axis of a finite line of charge   Instead of looking for the magnitude of the electric field adjacent to one of the ends of the finite line of charge, let's now examine the electric field along a perpendicular bisector at point P as shown in the diagram below. The charge present on a small segment of the rod, Δxi, can be expressed as Δqi which equals We do not have to be concerned with the horizontal, or x-components, of the electric field at point P since symmetry reveals that they will cancel. Consequently, all we have to do is add together all of the vertical components of the electric field at P and we will be done.   Considering this small section to represent one of hundreds or thousands of point changes comprising the rod, each Dqi has a vertical contribution to the electric field at point P can be represented by Adding together all of the contributions gives us the following expressions. Taking the limit as Dx approaches 0, we get the expression However, there is a problem with this integral. We cannot integrate r and  cos qwith respect to dx. We must standardize our variables. To achieve this, we are going to use two trigonometric substitutions: one for cos q and the other for tan q.   Referencing our initial charge diagram, we see that Substituting this for 1/r in our previous equation gives us Our expression is now slightly better since it no longer contains r and the constant a is easily removed from the integral whenever it is convenient. Our next step is to replace dx with dq. To do this will now use the trig function tan q. Isolating dx, gives us the expression Now we can write our final integral and evaluate. To finish we now need to return to our charge diagram and pay attention to assigning the limits to our integral. Looking from left to right, q sweeps from q1 to q2. Since we know that q1and q2 have the same magnitude (remember that P is on the perpendicular bisector of our finite line of charge)  we can write the relationship that q2 = -q1 Since q1 and q2 are equal, we can renamed our angle as just q without the need of any further subscripts.   But what about the length of the rod and its overall charge. How can we write an expression containing Q and L? To do this, we will use the following expression for sin q. Substituting gives us and we are finished.
 Case III: P along the axis of a charged ring   Now let's assume that our finite line of charge has been curved into a circle. What would be the electric field at the center of the circle? What would be the electric field anywhere along the circle's axis?   Once again, we state with a small charge segment and add up the total affects of all segments at point P. In the following diagram, the radius of the charged ring is a, and the distance along the axis to P is x. Once again, symmetry shows us that the y-components of the electric field will cancel, leaving only the x-components' contributions. This derivation will be much simpler than the previous linear bisector since the values for x, a, and r keep the same value for all charge segments, Dq, around the ring. In order to simplify this integral we will use the trig function cos q. Substituting yields and we are finished.
 Case IV: P at the center of a charged arc Each charge segment will contribute an electric field, DE, at point P. where Dq has been initially replaced with the expression lDs and Ds with a Dq. Notice that the electric field at P has both x- and y-components. That is, there is no symmetry to cancel either all of the x- or all of the y-components. We can write each of these components as As we pass to the limit as Dq goes to 0 and add up all of the contributions, we get the expressions: Thus our answer is and we are done.
 ADVANCED: P off the axis of a finite line of charge   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 (Ex) and vertical (Ey) 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. By referencing a table of integrals, we find that allowing us to integrate and calculate our final expression for Ey. We will now repeat our process and solve for Ex.   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 Ex becomes Some interesting consequences for this last derivation   If we place point O at , then Ex = 0 since the left since the Δ Ex vectors cancel in corresponding pairs. This would leave us with only net E = Ey. Suppose we view the bar from a very far distance; such that, and . Our expressions for Ex and Ey would reduce to  Related Documents