A rod of length L has a total charge Q distributed uniformly along its length .It is bent in shape of a semicircle.Find the magnitude of the electric field at the center of curvature of the semicircle?

Draw diagram and take 2 symmetric points and observe that the cos@ components cancel out., where @ is the angle the line joining the point to the centre makes with the base. Rest all is simple integration :)

Conjurer,please solve this one.rates assured. pleaseeee

E at the centre of curvature of a curve which subtends an angle theta at centre is 2kx/r .sin(theta/2)

this is 2kx/r for semicircle,where x is charge per unit length=Q/L,r is radius=L/pi,k=1/4pi

so,the ans is 2piKQ/L²=Q/2L²

bettr use  integration..... its methud applicable almost all times......

any dx element is taken let electric field due to it be as shown

now ........its component along horizontal direction gets cancelled out due to symmetry... only vertical component remains....charge density ( linear) on rod is lambda= Q/l  ... whre l is the length = 2 pie R..... whre r is radius of semicircle. so dQ = lambda (dl)............ and dl = R d (theta)

E due to dx elment at centre is .. KdQ / R^2...

vertical component =KQ/R^2( sin theta).... now this is integrated frm 0 to pie/2 as limits..              after integrating = KQ/R^2... as cos 0 =1 and cos 90= 0

This is for 1 half of semicircle and for other half

for total multiply by 2.........

ans. 2KQ pie ^2/ l^2    ............ whre l= length... maybe .. calculate for urself... bettr... mthud is same

Cheers!!!!!!

But answer is Q/2E₀L^2.please explain

oh!!!!  rit........ me so stupid....... i didnt gve importance to solving.......see ............

int. of ................. K dQ/ R^2 = K lambda dl / R^2

dl= Rd theta....... subst...  K lambda d theta/R...

this is electric field due toany element dx............ now integrating this frm theta = 0 - pie ........ we get......

K lambda pie / R........ put R= l/2 pie....and lambda = Q/l ....... and K= 1/4pie epsilon

to get required ans......... Q/2E₀L²

******or else u can use the exact formula given in earlier post..... integration though is safest and most app. methud

Cheers!!!!!