Question

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?

Answer 1

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.

Answer:

a)
`E = (mv_(0)^(2) y_(h) )/(0.5qL^(2) )`

b)
`E = (mv_(0)^(2) L )/(0.5qy_(h) ^(2) )`

Step-by-step 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 = (0.5qL^(2) E)/(mv_(0)^(2) )`

If the charge is to hit the target on the screen
`y = y_(h)`

`y_(h) = (0.5qL^(2) E)/(mv_(0)^(2) )`

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

`E = (mv_(0)^(2) y_(h) )/(0.5qL^(2) )`

b)

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 = (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) = (0.5qy_(h) ^(2) E)/(mv_(0)^(2) )`

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

`L = (0.5qy_(h) ^(2) E)/(mv_(0)^(2) )`

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

`E = (mv_(0)^(2) L )/(0.5qy_(h) ^(2) )`