Electric Potential and Kinetic Energy of a point chargeby pintsize131 Tags: electric field, electric potential, energy, point charge, voltage 

#1
Sep2811, 05:50 PM

P: 5

A uniform electric field has a magnitude 2.40 kV/m and points in the +x direction.
(a) What is the electric potential difference between x = 0.00 m plane and the x = 3.90 m plane? (b) A point particle that has a charge of +3.40 µC 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 = 3.90 m plane? (c) What is the kinetic energy of the particle when it arrives at the x = 3.90 m plane? (mJ) (d) Find the expression for the electric potential V(x) if its value is chosen to be zero at x = 0. (Use the following as necessary: x.) (kV) 2. Relevant equations (a)F=(kq_{1}q_{2})/r^{2} (b)E=(kq)/r^{2} (c)V=(kq)/r (d)EPE=U=(kq_{1}q_{2})/r 3. The attempt at a solution I figured out parts (a) and (b). (a) 9.36 kV (used equation c) (b) 31.82 mJ (used F=Eq, then F*r) (c) I got stuck here. My textbook said to "equate potential energy to kinetic energy." Does this mean that potential energy is the same as kinetic energy for the problem? (d)For this part, I don't understand the question. Any clarification? 



#2
Sep2911, 08:16 AM

P: 939

re (c):
The positive charge moves in the direction of the electric field i.e. in direction of decreasing electric potential. Hence as decrease in PE is transformed into KE. re (d): You are asked for the potential as a function of x. 


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