Electric force can be expressed as:

$\overline{){\mathbf{F}}{\mathbf{=}}{\mathbf{q}}{\mathbf{E}}}$

Newton's second law:

$\overline{){\mathbf{\Sigma}}{\mathbf{F}}{\mathbf{=}}{\mathbf{m}}{\mathbf{a}}}$

**1.**

The net electric force along the y-direction is:

F_{y} = qE

From Newton's second law, net acceleration in the y-direction is:

a_{y} = F/m = qE/m

A charge with mass m and charge q is emitted from the origin, (x,y)=(0,0). A large, flat screen is located at x=L. There is a target on the screen at y position y(h), where y(h) > 0. In this problem, you will examine two different ways that the charge might hit the target. Ignore gravity in this problem.

1.Assume that the charge is emitted with velocity v(0) 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, y(h), v(0), and L.

2.Now assume that the charge is emitted with velocity v(0) in the positive y direction. Between the origin and the screen, the charge travels through a constant electric field pointing in the positive x 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, y(h), v(0), and L.

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