Question

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.

Answer 1

E = - 2*E₀*sinθ*i

V = V = 2*E₀*sinθ /r

Step-by-step 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 )