An insulating rod is bent into the shape of a quarter-circle having total charge Q distributed uniformly over its length. At the origin, the magnitude of the total electric field due to this charge configuration is E0 and the potential is V0. A second insulating rod is added to the fourth quadrant to complete a semi-circle. This second quarter-circle has charge -Q uniformly distributed. In terms of E0 and V0, determine the total electric field vector and the total potential at the origin due to the semi-circle of charge.

E = - 2*E₀*sinθ*i

V = V = 2*E₀*sinθ /r

Explanation:

First situation: a quarter-circle rod produces at the origin

E₀      and  V₀

We know that    E  =  K *  Q/r      and

E  =  E₀           Q = L*λ           Q = 1/4 * 2*π*r*λ       Q = (1/2)*π*r*λ

where:

r is the radius of the circle which at the same time is the distance between the rod quarter of the circle and the origin.

λ is the longitudinal density of charge, and

k = 1/4*π*ε₀

Then    E₀ =  (1/2)* k*π*λ        and   V₀  = - E₀/Q         V₀  = - E₀/r

E₀  =  E₀ₓ  +  E₀y      E₀ₓ  = E₀*cosθ    E₀y = - E₀*sinθ  

θ the angle between E₀  and the x-axis

Second situation half of a circle ( first quarter +Q and fourth -Q )

In that configuration, the components in the x-axis cancel each other ( by symmetry) and the y-axis components of +Q  and -Q  are to be added then

E = Eₓ   +  Ey     but by inspection we find Eₓ  = 0 ( the components of the        

the electric field produced of the two quarters canceled each other)

and we have to add the two identical y-axis components of the two quarters therefore

E = 2*E₀y

each component of   E₀y = - E₀*sinθ    

Then the new electric field becomes

E = - 2*E₀*sinθ*i

The new V  = - (- 2*E₀*sinθ /r)

V = 2*E₀*sinθ /r

(Note remember that V is a scalar )

d eras and periods

area of focus