Educator

Electric Field of a Continuous Charge Distribution

• We know that the E-field produced by a point charge q at a point P a distance r away is (kq/r^2)*r_hat, where r_hat is a unit vector along the direction from q to P. How can we use this knowledge to find the E-field for a charge distribution?
• The answer is the following: Divide the charge into many, many small elements, each element being so small that it is essentially a point. Then, the E-field produced by each element dq is obtained by using the formula for the field of a point charge. The total E-field is obtained by adding all the fields produced by all the charge elements; in other words, the total E-field is the integral over the charge distribution of (kdq/r^2)*r_hat.
• Examples are given that show how to calculate the E-field produced by a charged rod, a charged ring, and a charged disk. In all these cases, the main issue is to set up the integral correctly, beyond which it becomes a problem in the calculus of integrations. In setting up the integral, we exploit the symmetry of the charge distribution to simplify the integral expression.