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?

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)

b)

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

If the charge is to hit the target on the screen

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

b)

The velocity of charge in the y direction = v₀

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

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

If the charge is to hit the target on the screen

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

i need with 1.3.2 investigate cycling o2 and co2 apex

we need to see it

electrical potential energy =0.072 j

explanation:

electrical potential energy is given by

u= q δv

q= charge=6 c

δv= potential difference and is given by

δv= e d

e= electric field

d= distance=5 mm=0.005 m

δv=2.4 (0.005)=0.012 v

now electric potential energy= u= q δv

u=6 (0.012)

u=0.072 j