Find the Electric Field

A uniformly charged insulating bar of length 14.0 cm is bent into a semicircle. The total charge on the bar is -7.50 μC.

What is the net electric field (in magnitude N/C and direction relative to the bar) at the point in the center of the semicircle?

Coulomb's constant is 8.99*10^9 N m^2 C^-2
The electromagnetic force is proportional to the inverse square of the distance

integrate that shit you raging fucking nigger faggot! the direction should be obvious from the symmetry of the semicircle. the field will go directly away from it. any field perpendicular to that will be canceled out by the other side of the semicircle. so you only have to work for one component of the field.

also, eat a dick and do you own homework.

also, saging the top thread

also, you're a nigger.

also, bahhh i dunno

fuck off

I haven't done something like this since mid last semester, but I assume this is an integration problem.

Consider a small segment of the charge bar dQ. dQ will equal the total charge Q times the fraction of the total arc it encompases, or dθ/π (dQ = dθ/π). Then consider the electric field from a similar infinitesimal segment of the bar, or dE = kdQ/|r|² * (-r^), where E and r are vectors. |r| = 0.14 m and -r^ = cosθ(-i) + sinθ(-j), so we get dE = -kdQ/0.14² * [cosθ(i) + sinθ(j)], and to simplify the RHS we get -0.14²dE/k = [cosθ(i) + sinθ(j)]*dQ.

Recall dQ= Q*dθ/π, so we get 0.14dE/kQ = cosθdθ/π(i) + sinθdθ/π(j)
We then integrate from θ = 0 to θ = π, so...
-0.14²E/kQ = [sin(π)/π - sin(0)/π](i) + [-cos(π)/π + cos(0)/π](j)
-0.14²E/kQ = [0/π - 0/π](i) + 1/π * [1+1](j)
-0.14²E/kQ = 0(i) + 2/π (j)

E = -2kQ/πr² (j) => E = -2(8.99*10^9)(-7.50*10^-6)/(0.14²π)(j), which will be positive, meaning the positive test charge in the middle of the semicircle would be attracted straight towards the point of the circle directly 'above' it.