Equipotential surface A has a potential of 5650 V, while equipotential surface B has a potential of 7850 V. A particle has a mass of 6.90 10-2 kg and a charge of +5.35 10-5 C. The particle has a speed of 2.00 m/s on surface A. A nonconservative outside force is applied to the particle, and it moves to surface B, arriving there with a speed of 3 m/s. How much work is done by the outside force in moving the particle from A to B?

Explanation:

Formula for the change in potential energy from point A to B is as follows.

           P.E =

Putting the given values into the above formula as follows.

          P.E =

                 =

                 = -0.11 J

Now, we will calculate the change in kinetic energy as follows.

            K.E =

                  =

                  = 0.1035 J

Therefore, supplied difference by the outside force is calculated as follows.

           0.1035 J - (-0.11) J

          = 0.2135 J

Thus, we can conclude that work is done by the outside force in moving the particle from A to B is 0.2135 J.

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think its the third option