How to Find Electric Field at Semicircle Center Without Calculus?

[PhysicsNewb]

Hi, I don't want help solving this problem, just a method to complete it without using calculus. Here it is:

A thin glass rod is bent into a semicircle of radius r. A charge +q is uniformly distributed along the upper half, and a charge -q is uniformly distributed along the lower half. Find the magnitude and direction of the electric field E at the center of the semicircle. (which is the center of a normal circle if one existed in this problem)

The key for me is, how to solve without using calculus, if that is possible. Thanks.

 

[Andrew Mason]

Hi, I don't want help solving this problem, just a method to complete it without using calculus. Here it is:

A thin glass rod is bent into a semicircle of radius r. A charge +q is uniformly distributed along the upper half, and a charge -q is uniformly distributed along the lower half. Find the magnitude and direction of the electric field E at the center of the semicircle. (which is the center of a normal circle if one existed in this problem)

The key for me is, how to solve without using calculus, if that is possible. Thanks.

Use Gauss' law.
AM

 

[wais]

One method for solving this problem without using calculus would be to use the concept of symmetry. Since the charge distribution is uniform along both halves of the semicircle, we can use the principle of superposition to break the problem into smaller, simpler parts.

First, we can consider the electric field due to the positive charge distribution on the upper half of the semicircle. This can be calculated using the formula for the electric field of a line of charge, E = kλ/r, where k is the Coulomb constant, λ is the linear charge density (in this case, q/πr), and r is the distance from the center of the semicircle to the point where we want to find the electric field.

Next, we can consider the electric field due to the negative charge distribution on the lower half of the semicircle. This will also be calculated using the same formula, but with a negative linear charge density of -q/πr.

Finally, we can use the principle of superposition to add the two electric fields together, since they are acting in the same direction (towards the center of the semicircle). The resulting electric field at the center of the semicircle will have a magnitude equal to the sum of the individual electric fields, and the direction will be towards the center of the semicircle.

This method does not require calculus, but it does rely on the concept of symmetry and the application of the principle of superposition. It may not be as efficient as using calculus, but it can still provide a solution to the problem.