Electric field in the center of a circle?

Question

Hi,

I am stumped on this question and wish to get some guidance:

You are helping to design an electron microscope. A new device to position the electron beam is a charged circle of conducting material, divided into two half circles separated by a thin insulator; one half can have a + charge and the other a - charge. The electron beam will go through the center of the circle, and your job is to calculate the electric field in the center of the circle as a function of the amount of positive charge on one half circle, the amount of negative charge on the other, and the radius of the circle. Use integration.

I am confused where to begin. I am guessing I will have to use E = (kq) / r^2 somewhere. Any ideas?


Thank you!

Answer 1

Use angle as the variable of integration, and decompose the electric field vector between the center and any infinitesimal segment of the circle into x and y components. This is easy, since it's just Sin(a) da and Cos(a) da (while letting r = 1 to simplify things, it doesn't matter since it's a constant). Integrating those functions, we have -Cos(a) and Sin(a), for a = 0 to pi (For each half circle), or 0 and 2, which makes sense since the x component electric field should cancel out by symmetry. The same is true on the other side, but the direction of electric field is the same (attractive + repulsive). The full equation for the electric field strength should be:

E = k ((2q1) +(2q2)) / r^2

where k is the constant, q1 is charge of the + ring, q2 is the charge on the - ring, and r is the radius of the circle.

Answer 2

i could agree that the sector on the midsection could be 0 with 4, or 6 costs. i'm uncertain that 5 costs could be completely balanced. properly, i assume so. A inflexible, weightless disk with 5 weights on the circumference could desire to be balanced. that's an equivalent subject. be conscious the transformation of a topic i come across demanding into an equivalent subject i come across trouble-free. in case you are able to study this trick, you will start to think of like a physicist. As Feynman says, "the comparable equations have the comparable strategies."