As we know that electrostatic force between two charges is given as
here we know that electrostatic repulsion force is balanced by the gravitational force between them
so here force of attraction due to gravitation is given as
here we can assume that both will have equal charge of magnitude "q"
now we have
now we have
Answer:
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
