Question

A particle with charge +4.20 nC is in a uniform electric field E directed to the left. The charge is released from rest and moves to the left; after it has moved 6.00 cm, its kinetic energy is+2.20 Times 10^-6 J. What is the work done by the electric force? Express your answer with the appropriate units. What is the potential of the starting point with respect to the end point? Express your answer with the appropriate units. What is the magnitude of E'?

Answer 1

Answer:

The  work done by the electric force is = $+2.20*10^{-6} J$

The potential of the starting point with respect to the end point is    $= 523.8 V$

The magnitude of E $= 8730 \ N/C$

Explanation:

Looking at this we can see that the work done by the electric force is equal to the kinetic energy this is because on the electric field is uniform and that the work done resulted in motion.

Mathematically

          $\Delta V q = W$

Where  $\Delta V$ is the potential difference

              q is the charge whose value i given as  = $4.20 *10 ^{-9} C$

               W is the workdone whose value is obtained as =  $+2.20*10^{-6} J$

  Substituting

                     $\Delta V = \frac{W}{q}$

                             $=\frac{2.2*10^{-6}}{4,20*10^{-9}}$

                               $= 523.8 V$

Mathematically  

                       $\Delta V = Ed$

        Where  E is the electric field strength

                       d is the distanced moved and it is given as 6.00 cm = 0.06 m

       Making E the subject of the formula

                    $E = \frac{\Delta V}{d}$

                      $E = \frac{523.8}{0.06} = 8730 \ N/C$