Find the electric field intensity at the center of

1.A quarter ring   2.semi-circular ring         3.3/4 th of the ring        4.a ring

; that is, . The net charge represented by the entire circumference of length of the semicircle could then be expressed as Q =

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 Ds_i along the semicircle. The charge present will be Dq_i where

Dq_i = lDs_i

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

Note that we must deal with both the horizontal (E_x) and vertical (E_y) 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, E_iy.

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

Remember that

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 (E_x) and vertical (E_y) 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:

Since

An interesting result occurs when

The value of E returns to that of a point charge.