Linear charge density?

Question

Hi,

A plastic rod with linear charge density lambda is bent into a quarter circle. The quarter circle is placed in the 3rd quadrant in a way that if 3 more quarter circles were placed on the plane, the would form a whole circle (one quarter circle in each quadrant). The plastic rod has a positive charge all over. Find the electric field at the origin.

This might be hard to picture so I tried my best in paint to represent what is going on: http://i137.photobucket.com/albums/q208/infinitbelt/Figure2649.jpg


QUESTIONS:

a. Write expressions for x and y components of the electric field at the origin due to a small piece of charge at angle theta.

b. Write, but do not evaluate, definite integrals for the x and y components of the net electric field at the origin.

c. Evaluate the integrals and write Enet in component form.


Any help would be greatly appreciated!

Thanks!

Answer 1

a. Elementary charge dq due to a small section of this rod can be expressed as lambda*R*d(theta). Radial component of electric field at the origin due to this elementary charge is dE = dq/R.
X and Y components can be expressed from this using trig functions. I hope it is correct and I hope it gives you some idea.

Answer 2

variables: T=angle theta : dT=delta theta/change of theta : R=radius : q=charge : L=lambda : k=constant


Electric Field (E.F.) strength is expressed as:
E=k( q/R^2 ) where q=charge

Consider small element of the quarter circle with angle theta(T), it has a charge of deltatheta(dT). The E.F. strength due to this piece is dE.
dE=k( dq/R^2 )

recall:
x=R(T)
dx=R(dT)

Charge expressed as linear charge density: (L=lambda)
dq=L(dx)=LR(dT)

The length of the element x is in terms of angle and:
dE=[kLR(dT)]/R^2 = [kL(dT)]/R

Now for components:
X-component of E.F. due to small piece is:
dEx=dEcosT
dEx=kL(dT)/R cosT
dEx=kL/R cosT(dT)
Ex=integral[0 to pi]{ kL/R cosT(dT) }
...pull out constants...
Ex=kL/R integral[0 to pi]{ cosT(dT) }
Ex=kL/R sinT [0 to pi] = kL/R *(1)
Ex=kL/R

Y-component of E.F. due to small piece is:
dEy=dEsinT
dEy=kL(dT)/R sinT
dEy=kL/R sinT(dT)
Ey=integral[0 to pi/2]{ kL/R sinT(dT) }
Ey=kL/R integral[0 to pi/2]{ sinT(dT) }
Ey=kL/r(-cosT) [0 to pi/2] = kL/R *(1)
Ey=kL/R

Thus E.F. net:
E=(kL/R) i + (kL/R) j or E=