A uniform electric Field of strength E exists in a region an electron of mass m enter at a point A perpendicular to x axis with velocity v. If moves through a electric Field exist at point B . The component of velocity at B is shown. (a) Find the Electric field (b) Find velocity at B (c) Find work done?

Question

Answer

Electric Field Calculation:
- The force experienced by the electron in the electric field is given by F = qE, where q is the charge of the electron.
- The force is also related to the acceleration of the electron, F = ma.
- Equating the two equations, we get qE = ma. Since q = -e (charge of an electron) and a = dv/dt, we have -eE = m(dv/dt).
- Solving this differential equation, we get v = -eEt/m + constant. Since the electron enters at point A perpendicular to x axis with velocity v, the initial velocity v0 = 0.
- Substituting the values, we get v = -eEt/m.

Velocity Calculation at B:
- At point B, the component of velocity along the x-axis is given as vx = -eEt/m.
- The velocity at point B is the x-component of velocity, so the velocity at B is vx = -eEt/m.

Work Done Calculation:
- The work done by the electric field on the electron is given by W = qEd, where d is the displacement of the electron.
- Since the electron moves perpendicular to the electric field from point A to point B, the work done is W = qE(dB - dA).
- Substituting the values, we get W = -eE(dB - dA).
- The work done can also be calculated using the change in kinetic energy, W = ΔKE. Since the initial velocity v0 = 0, KE at point A is 0. The final kinetic energy at point B is 1/2mvB^2.
- Therefore, W = 1/2mvB^2 - 0 = 1/2mvB^2.
In conclusion, the electric field strength E, velocity at point B, and work done on the electron can be calculated using the given information and the equations of motion for charged particles in electric fields.

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