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Post by Daniel Brook on February 21, 2011

The Y-component aspect of the continuous charge distribution at the perpendicular bisector was really hard to understand. All of the trigonometry became really confusing. Either a different method of solving would have been nice or a more in-depth explanation of the problem could have helped.

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Last reply by: Kyle Kosic
Mon Apr 4, 2011 8:42 PM

Post by Kyle Kosic on April 4, 2011

When he has 1/a^2 * sin(theta), couldn't he replace sin(theta) with x/a?

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.

Electric Field of a Continuous Charge Distribution

Lecture Slides are screen-captured images of important points in the lecture. Students can download and print out these lecture slide images to do practice problems as well as take notes while watching the lecture.

AP Physics C: Electricity and Magnetism