Work and electric field

[silver_gry]

Homework Statement

Equipotential surface A has a potential of 5650V, while equipotential surface B has a potential of 7850V. A particle has a mass of 5.00*10^-2kg and a charge of +4.00*10^-5. The particle has a speed of 2.00 m/s on the surface A. An outside force is applied to the particle and it moves to surface B, arriving there with a speed of 3.00 m/s. How much work is done by the outside force in moving the particle from A to B?

Homework Equations

The Attempt at a Solution

Can someone please help me with this problem? I really need help with how to start this problem and what equations to use! Please help.

 

[LowlyPion]

What is the equation for relating Work and equipotential surfaces?

What changes in kinetic energy have occurred?

 

[nlightner]

The work done by an outside force on a charged particle can be calculated using the equation W = qΔV, where W is the work done, q is the charge of the particle, and ΔV is the change in potential between the two surfaces.

In this case, the particle has a charge of +4.00*10^-5 and the potential difference between the two surfaces is 7850V - 5650V = 2200V.

Plugging these values into the equation, we get W = (+4.00*10^-5)(2200) = 0.088 J.

Therefore, the outside force did 0.088 J of work in moving the particle from surface A to B.