Question

two plastic curved rods, one of charge q and the other of charge -q. which form a semicircle of radius r=8cm in xy. If q=3.8pC, what would be the magnitud and direction of the electric force E produced in P,in the center.

Answer 1

let us find out the electric field due to a charged arc at the center when it sustend 2t angle at the center and total charge is q

let uniform chare per uit length L = q2t/2R. = qt/R

let us consider a small element of thickness dl. Let this dl make a small angle at the origin dx and be at the end of the line which makes angle x with the X axis.
Hence the dl = Rdx
The charge for this element will be Rdx L.
The field at the origin will be K RL dx / R^2 = K *(L/R)* dx
This vector is away from the charged element along the line which makes an angle x. Let us resolve this vector along negative Y axis as K *(L/R)* dx* sin x and K *(L/R)* dx* cos x along negative X axis.
BY symmetry all the sine components will get cancelled where as we have to add the cosine components.
Hence the resultant field will be 2*K *(L/R)* Integ dx cos x
(by symmetry) within the limits 0 to t.
This gives 2*K *(L/R)* [- sin x] with limit 0 to t.
Hence the net field = 2*K *(L/R)* sint.

acoording to ur question one quater is neatively charged and one is positively charged

so one part will produce electric field towards itself

adding the two will give a electric field as 2Ecos(45⁰) (towards the negative end of semicircle) as angle between two quater's center is 45

E= 2*2*K *(L/R)* sint*cos45. here 2t = 45⁰ L = 3.8pC*(/8)/0.08 = 3.8 * 10^-10 C/m

E= 4*9*10^9*(3.8*10^-10/0.08)sin22.5 cos45 = 46.17 N/c