A circular arc has a uniform linear charge density of 4 nC/m. 287 ◦ 2.5 m x y What is the magnitude of the electric field at the center of the circle along which the arc lies? The value of the Coulomb constant is 8.98755 × 109 N · m2 /C 2 . Answer in units of N/C.

294.0724 N/C

Explanation:

From the question, we are given the following parameters: linear charge density,α = 4 nC/m = 4 × 10^9 C/m, radius, r= 2.5 m, the angle in the counter clockwise direction from the positive x-axis= θ= 287° and the Coulomb constant,Kc= 8.98755 × 109 N · m2 /C 2.

So, we can solve this question by using the formula below;

==>E(x direction)= - (Kc ×α )/ r × {sin 287° - sin 0°} i.

==>> - (8.98755 × 10^9 × 4 × 10^-9)/2.5 × { −0.9563047559630354

- 0} I.

==> -14.4 × [−0.9563047559630354].

==> 13.8 N/C i.

E( y direction)= - (Kc ×α )/ r × ( cos 0° - cos 287°) j.

==> -14.4 × (1 - [ 0.29237170472273672]).

==>( -14.4 × 0.7076) j.

==> - 10.18944 n/C j.

==>E(x direction)^2+ E( y direction)^2.

==> (13.8^2 + (- 10.18^2) N/C.

==> 190.44 N/C. + 103.6324 N/C.

= 294.0724 N/C

uranium fissions/breaks up giving out energy and neutrons