Thank you Uniform Electric Field Given a uniform electric field of magnitude 30.

Question

Thank you

Uniform Electric Field Given a uniform electric field of magnitude 30.0 N/C pointing in the positive x-direction and a positive charge q = 2.5 C The charge moves from x = 0.0 m to x = 2.0 m, how mu Ch work is done by the electric field on the charge? What is the change in potential energy as the charge moves from x = 0.0 m to 2.0 m? What is the speed of the positive charge when it reaches 2.0 m? The mass of the charge = 3.0 kg. What is the change in potential as the positive charge moves from x = 0.0 m to x = 2.0 m? Using your answer from part d), what is the change in potential energy as the charge moves from x = 0.0 m to x = 2.0 m?

Explanation / Answer

a)work=force*distance=30*q*x=150

b)work done by electric field=chane in potential enrgy.So answer=-150J

C)as energy is conserved change in potential energy=change in kinetic energy

so 150=0.5*3*v^2 so v=10m/sec

d)change in potential=change in potential energy/charge=-75

e)-150

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