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Ugo [173]
2 years ago
5

A +2e charge is at the point (-1,0) mm in the x,y plane. A –e charge is at the point (0,1) mm. What is the electric field at the

origin? What would be the force on a +3e charge at this point? Give the x and y components for both the field and force
Physics
1 answer:
Gennadij [26K]2 years ago
4 0

Answer:

Let I and j be the unit vector along x and y axis respectively.

Electric field at origin is given by

E= kq1/r1^2 i + kq2/r2^2j

= 9*10^9*1.6*10^-19*/10^-6*(2i+ j)

= (2.88i + 1.44j)*10^-3 N/C

Force on charge= qE= 3*10^-19*1.6*(2.88i +1. 44 j) *10^-3

F= (1.382 i + 0.691 j) *10^-21

 

 Goodluck

Explanation:

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Answer:

150 km/hr²

Explanation:

The data for the motion of the train includes;

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The acceleration, 'a', is the rate of change of velocity with time;

a = \dfrac{\Delta v}{\Delta t} = \dfrac{v - u}{\Delta t}

Therefore, we have;

a = (75 km/hr - 0 km/hr)/(0.5 hr) = 150 km/hr²

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<h2>Answer:</h2>

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