Question

A charge of 4 × 10^{- 9}c is distributed uniformly over the circumference of a conducting ring of radius 0.3 m. Calculate the field intensity at a point on the axis of the ring at 0.4 m from its center and also at the center?

पर्याय

• 112 N/C, 2 N/C 

• 112 N/C, 3 N/C 

• 115.2 N/C, Zero 

• 113.2 N/C, Zero

Answer 1

115.2 N/C, Zero

Explanation:

q = 4 × 10^{- 9}C

r = 0.3 m, d = 0.4 m

E = ?

As we know:

E = `"qd"/(4piepsilon_0 ("d"^2 + "r"^2)^(3//2))`

E = `(9 xx 10^9 xx 4xx10^-9 xx 0.4)/(0.4^2 + 0.3^2)^(3/2)`

E = `14.4/(0.5)^3 = 115.2` N/C

At the center d = 0, E = 0