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Kkampudiaa
26.08.2020 •
Physics
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?
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Ответ:
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
Ответ:
January 24th
Explanation:
It is about 21 days, so add 21 and 3, 24, so on January 24th