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 5.40 10-2 kg and a charge of 5.10 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

0.247 J = 247 mJ

Explanation:

From the principle of conservation of energy, the workdone by the applied force, W = kinetic energy change + electric potential energy change.

So, W = ΔK + ΔU = 1/2m (v₂² - v₁²) + q (V₂ - V₁) where m = mass of particle = 5.4 * 10⁻² kg, q = charge of particle = 5.10 * 10⁻⁵ C, v₁ = initial speed of particle = 2.00 m/s, v₂ = final speed of particle = 3.00 m/s, V₁ = potential at surface A = 5650 V, V₂ = potential at surface B = 7850 V.

So, W = ΔK + ΔU = 1/2m (v₂² - v₁²) + q (V₂ - V₁)

= 1/2 * 5.4 * 10⁻²kg * ((3m/s) ² - (2 m/s) ²) + 5.10 * 10⁻⁵ C (7850 - 5650)

= 0.135 J + 0.11220 J

= 0.2472 J

≅ 0.247 J = 247 mJ