What is the electric field at a location vector b = <-0.2, -0.4, 0> m, due to a particle with charge +3 nC located at the origin?

E = 134.85 N/C

Explanation:

We are given;

Electric field location; b = <-0.2, -0.4, 0> m

Charge: q = 3 nc = 3 × 10^(-9) C

From the location given, we have total distance;

r = √((-0.2²) + (-0.4)² + (0²))

r = √0.2

Formula for Electric field is;

E = kq/r²

where;

k is a constant with a value of 8.99 x 10^(9) N.m²/C²

q is charge on the proton particle with a costant value = 1.6 × 10^(-19) C

r is the distance

Plugging in the relevant values, we have;

E = (8.99 x 10^(9) × 3 × 10^(-9))/(√0.2)²

E = 134.85 N/C

answer: 31, 5556. 952 seconds