A charge q of mass m starts with a velocity vo in the +x direction. It passes th

Question

A charge q of mass m starts with a velocity vo in the +x direction. It passes the origin when an electric field of strength E pointing in the -x direction is switched on. (a) How far would the charge go before it comes to a stop? (b) Now let us say that instead of a constant E-field on the -x direction, that another charge q lies ahead of our moving charge at a distance from the origin along the x-axis. Assuming that this charge is stationary, how far from the origin would it have to be to stop our moving charge in the same spot as part (a)?

Explanation / Answer

A charge q of mass m starts with a velocity vo in the +x direction. It passes the origin when an electric field of strength E pointing in the -x direction is switched on.

(a) How far would the charge go before it comes to a stop?

force=Eq

let v be the velocity of the particle acquired due to the electric field.

let the stopping distance be x

so 1/2m v*v=Eq*x

so x=[1/2m v*v]/Eq

(b) Now let us say that instead of a constant E-field on the -x direction, that another charge q lies ahead of our moving charge at a distance from the origin along the x-axis. Assuming that this charge is stationary, how far from the origin would it have to be to stop our moving charge in the same spot as part (a)?

Eq=9*109q*q/x*x

so x = [9*109q*q/Eq]

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