A half circle arc with its center at the origin has a linear charge density of λ = 12 nC/m along the arc. The radius of the arc is R = 3 cm. (a) What are is the electric field (magnitude and direction) at the origin? (b) What is the electric potential, V , at the origin?

Explanation:

Given that,

Charge density is  λ = 12 nC/m

And radius 3cm

r=0.03m.

The charge density of a along a circular arc is given as

λ= Q/πr

Then, Q=πrλ

Q=π×0.03×12×10^-9

Q=1.131×10^-9 C

Q=1.131 nC

Then, electric field along x axis is symmetrical and if cancels out

Now, Ey is in the negative direction

Electric field is given as,

Ey=-2kQ/πr²

K is constant =9×10^9Nm²/C²

Ey=-2×9×10^9×1.131×10^-9/(π ×0.03²)

Ey=7200 N/C.

The direction is negative direction of y axis, check attachment for diagram.

b. Electric potential at the origin is given as

V=Ed

d=r=0.03

V=7200×0.03

V=216V

this means that the temperature of both substances will be equal

the bal will not change its velocity.