A particle leaves the origin with a speed of 3.6 106 m/s at 34 degrees to the positive x axis. It moves in a uniform electric field directed along positive y axis. Find Ey such that the particle will cross the x axis at x

E = - 4556.18 N/m

Explanation:

Given data

u = 3.6*10^6 m/sec

angle = 34°

distance x = 1.5 cm = 1.5*10^-2 m (This data has been assumed not given in

Question)

from the projectile motion the horizontal distance traveled by electron is

x = u*cosA*t

⇒t = x / (u*cos A)

We also know that force in an electric field is given as

F = qE

q = charge, E = strength of electric field

By newton 2nd law of motion

ma = qE

⇒a = qE/m

Also, y = u*sinA*t - 0.5*a*t^2

⇒y = u*sinA*t - 0.5 * (qE/m) * t^2

if y = 0 then

⇒t = 2mu*sinA / (qE) = x / (u*cosA)

Also, E = 2mu^2*sinA*cosA / (x*q)

Now plugging the values we get

E = 2*9.1*10^{-31}*3.6^2*10^{12} * (sin34°) * (cos34°) / (1.5*10^{-2} * (-1.6) * 10^{-19})

E = - 4556.18 N/m