Question

A particle with mass m and charge q is emitted from the origin, (x,y)=(0,0). A large, flat fluorescent screen is located at x=L. There is a target on the screen at y position yh, where yh>0. Ignore gravity in this problem.

Part A Suppose the charge is emitted with velocity v0 in the positive x direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive y direction. What should the magnitude E of the electric field be if the charge is to hit the target on the screen? Express your answer in terms of m, q, yh, v0, and L.

Answer 1

Solution,

Let the horizontal distance be L and horizontal velocity be v0, then

Time, t = L/v0

Using equation,

yh = vt + 0.5 at^2 ...... (1)

Force, F = qE = ma

Acceleration, a = F/m = qE/m

yh = 0 + 0.5 (qE/m) (L/v0)^2

E = (2my_h/q) (v₀/L)²

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