A uniform electric field has a magnitude 3.20 Kv/m and points in the -x directio

Question

A uniform electric field has a magnitude 3.20 Kv/m and points in the -x direction .

1- what is the electric potential difference between x-0.00 m plane and the x-4.60 m plane?

2- a point particle that has a charge of -8.10 muC is released from rest at the origin. what is the change in the electric potential energy of the particle as it travels from the x-0.00m plane to the x-4.60 m plane?

3-what is the kinetic energy of the particle when it arrives at the x-4.60 m plane?

4-find the expression for the electric potential l'(x) if its value is chosen to be zero at x-0

Explanation / Answer

1.V=Ed=-4.6*3200=14720 Volts

charge=-8.1*10^(-6) C

2.Energy=e*V=-0.119 J

3.KE=0.119 J

4.l'(x)=3200x

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