(A) Suppose you need to calculate the electric field at point P located along the axis of a uniformly charged semicircle. Let the charge distribution per unit length along the semicircle be represented by l; that is, . The net charge represented by the entire circumference of length of the semicircle could then be expressed as Q = l(pa).
Let's first combine F = qE and Coulomb's Law to derive an expression for E.
For the semicircle of charge shown below
examine a small section Dsi along the semicircle. The charge present will be Dqi where
The electric field contribution at P by this section would be represented by
Note that we must deal with both the horizontal (Ex) and vertical (Ey) components of the electric field at P. Due to symmetry, the horizontal components will cancel, and the net electric field can be calculated by summing up the contributions by the vertical components, Eiy.
Taking the limit as Ds approaches 0, we can express Ey as
Unfortunately this leaves us with an expression involving two variables: s and q.
In our case, since r = a the equation becomes
Placing the correct limits on the integral and evaluating gives
Replacing we obtain the final expression for the net E at point P.
(B) Suppose you are now asked to calculate the electric field at point P located a distance b from the center of the center of a uniformly charged ring with a charge per unit length, . Notice in the following diagram that we must deal with both the horizontal (Ex) and vertical (Ey) components of the electric field at P.
Using vertical angles and right triangle trigonometry, we can calculate that
Due to symmetry, all of the x-components will cancel, allowing us to sum up the y-components to determine the net electric field at P.
Taking the limit as Ds approaches 0, we get that
Unfortunately this leaves us with an expression involving three variables: s, r, and q. Since our differentiable is ds, we need to replace r and q with equivalent expressions involving only s. We can make this happen by noticing the following relationships:
Substituting for r and cos q, we get
Since gives us the circumference of the ring, we have:
An interesting result occurs when
The value of E returns to that of a point charge.
|Copyright © 1997-2013
Catharine H. Colwell
All rights reserved.
Mainland High School
Daytona Beach, FL 32114