At t = 0 a very small object with mass 0.400 mg and charge +9.00 µC is traveling at 125 m/s in the ?x-direction. The charge is moving in a uniform electric field that is in the +y-direction and that has magnitude E = 895 N/C. The gravitational force on the particle can be neglected. How far is the particle from the origin at t = 7.00 ms?

Solution 74P Step 1 of 4: We have to find the force on the charged particle due to the electric field. Then using Newton’s second law we can find acceleration and the constant acceleration kinematics formulas to find the components of the distance it moves. The electric force on a charged particle along xc axis is F =xqE x According to Newtons secnd law , The force along x axis F = ma x x 1 2 Acceleration is constant in x direction so, we can use kinematic eqn, x x = v t + a0 0x 2 Step 2 of 4: The only non zero acceleration in the y direction so a = 0 x F y F =yqE y and F = my y so, a y m The electric force along y axis , 6 3 F =y(9 × 10 C)(895 N/C) = 8.055 × 10 N