A uniform electric field has a magnitude 2.00 kV/m points in the +x direction. (

Question

A uniform electric field has a magnitude 2.00 kV/m points in the +x direction. (a) What is the electric potential difference between the .v = 0.00 m plane and the x - 4.00 m plane? (b) A point particle that has a charge of +3.00 mu C 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 - 0.00 m plane to the x - 4.00 m plane? (c) What is the kinetic energy of the particle when it arrives at the x = 4.00 m plane? (d) Find the expression for the electric potential V(x) if its value is chosen to be zero at x = 0.

Explanation / Answer

Given

   uniform electric field E = 2000 V/m

  
a) we know that electric potential difference is V = E*d

   V = 2000*(4-0) = 8000 V

b) by the definition of potential

   the work done per unit +ve test charge to bring it from infinity to a point inside the electric field.

that is V = W/q ==> V = U /q ==> U = V*q = 8000*3*10^-6 = 24*10^-3 J

c)   kinetic energy is , work done = change in k.e

               W = 24*10^-3 J = (k2-k1)

               k2-k1 = 24*10^-3 J

d) electric potential is V = E*d or v(x) = kq/r

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