Question

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?

Answer 1

Explanation:

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

           P.E = $(V_{A} - V_{B}) \times q$

Putting the given values into the above formula as follows.

          P.E = $(V_{A} - V_{B}) \times q$

                 = $(5650 - 7850) \times 5 \times 10^{-5}$

                 = -0.11 J

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

            K.E = $0.5 \times m \times (v^{2}_{B} - v^{2}_{A})$

                  = $0.5 \times 6.90 \times 10^{-2} \times (2^{2} - 1^{2})$

                  = 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.