Question

The figure below shows a small, charged bead, with a charge of q +45.0 nC, that moves a distance of d 0.163 m from point A to point B in the presence of a uniform electric field E of magnitude 290 N/C, pointing right B (a) What is the magnitude (in N) and direction of the electric force on the bead? 400 magnitude N Salect- direction (b) What is the work (in J) done on the bead by the electric force as it moves from A to B? 40 (c) What is the change of the electric potential energy (in J) as the bead moves from A to B? (The system consists of the bead and all its surroundings.) PEB PE = J (d) What is the potential difference (in V) between A and B? VVa4

Answer 1

(a) The force on a charged particle of charge 'q' in the electric field of strength 'E' is given by

                                  

F = gE

So, for the given problem, the force on the bead is

                                     in the rightward direction.

F = 45 x 109 x 290 = 13.05 x 10-6 N

(b) The work-done is

                                      

W = F.AS = 13,05 x 10-6 x 0.163 = 2.13 x 10 J

(c) The electric potential energy can be calculated as

                                    

E.dr AU =

Or from the work-energy theorem

                                    $\Delta U = -W \,\, \text{(Workdone)}$

For the given problem,

                                    $\Delta U = -2.13 \times 10^{-6}\,\, J$

(d) The electric potential can be calculated as

                                  

AU AV

So, for the bead

                                   $\Delta V= \frac{ -2.13 \times 10^{-6}}{45 \times 10^{-9}} =-47.27\,\, volts$