Physics, 15.04.2020 00:55 AdoNice

Now assume that the charge is emitted with velocity v0 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?

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Answer from: MarishaTucker

Complete Question:

An charge with mass mand charge qis 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 positiony_h, where y_{\rm h} /> 0 . In this problem, you will examine two differentways that the charge might hit the target. Ignore gravity in thisproblem.

a.)

Assume that the charge is emitted withvelocity v_0 in the positive x direction. Between the originand the screen, the charge travels through a constant electricfield pointing in the positive y direction. What shouldthe magnitude Eof the electric field be if the charge is to hit the target on thescreen?

Express your answer in terms ofm,q,y_h, v_0, and L.

b.)

Now assume that the charge is emitted withvelocity v_0 in the positive y direction. Between the originand the screen, the charge travels through a constant electricfield pointing in the positive x direction. What shouldthe magnitude E of the electric field be if the charge is to hit the target on thescreen?

Express your answer in terms ofm,q,y_h, v_0, and L.

a) $E = \frac{mv_{0}^{2} y_{h} }{0.5qL^{2} }$

b) $E = \frac{mv_{0}^{2} L }{0.5qy_{h} ^{2} }$

Explanation:

The velocity of charge in the x direction = v₀

The time required by the charge to hit the screen at a distance L,

t = L/v₀

The force on the charge, F = qE

F = ma

Equating the two relations for force

ma = qE

acceleration of the charge, a = qE/m

The initial velocity of the charge along the y-direction is o, the vertical distance covered by the charge is:

y = 0.5at²

substituting the relations for a and t

y = 0.5* (qE/m)*(L/v₀)²

$y = \frac{0.5qL^{2} E}{mv_{0}^{2} }$

If the charge is to hit the target on the screen $y = y_{h}$

$y_{h} = \frac{0.5qL^{2} E}{mv_{0}^{2} }$

Making E the subject of the formula, the magnitude of the electric field is:

$E = \frac{mv_{0}^{2} y_{h} }{0.5qL^{2} }$

The velocity of charge in the y direction = v₀

The time required by the charge to hit the screen at a height $y_{h}$ ,

$t = \frac{y_{h} }{v_{0} }$

The force on the charge, F = qE

F = ma

Equating the two relations for force

ma = qE

acceleration of the charge, a = qE/m

The initial velocity of the charge along the x-direction is o, the horizontal distance covered by the charge is:

x(t) = 0.5at²

substituting the relations for a and t

$x(t) = 0.5(qE/m)(y_{h}/v_{0} ) ^{2}$

$x(t) = \frac{0.5qy_{h} ^{2} E}{mv_{0}^{2} }$

If the charge is to hit the target on the screen x(t) = L

$L = \frac{0.5qy_{h} ^{2} E}{mv_{0}^{2} }$

Making E the subject of the formula, the magnitude of the electric field is:

$E = \frac{mv_{0}^{2} L }{0.5qy_{h} ^{2} }$

Answer from: Quest

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Answer from: Quest

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Now assume that the charge is emitted with velocity v0 in the positive y direction. Between the orig...