A particle charge of 7.6μC is in a uniform electric field directred to the left. Another force, in addition to the electric force, acts on the particle so that when it is released from rest it moves to the right. After it has moved 8cm, the additional force has done 6.5x10^-5J of work and the particle has 4.35x10^-5 J of kinetic energy.
- Calculate work done by Electric Field
- Find the potential of the starting point with respect to the end point
- Find the Magnitude of Electric Field
- W(F) = 6.5x10^-5J
- W(E) = unknown
- Kb = 4.35x10^-5J
- Ka = 0
- Ua = unknown
- Ub = unknown
- q = 7.6μC
- r = 0.08m
W = Ua - Ub = Kb - Ka
K = 0.5MV^2
U = V = 1/4πε * q/r
E = 1/4πε * q/r^3 * r
The Attempt at a Solution
- Using the work-energy theorem, I can say that the work done by the electric field is 4.35x10^-5J - 6.5x10^-5J.
- The potential between 2 points equals the amount of work that would be required to move a unit charge between those points: So I Believe I should just plug in variables for V.
- Same as 2. I'll plug in for E, using the -i unit vector to represent the direction.