Question

A particle leaves the origin with a speed of 2.1 times 106 m/s at 30 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 = 1.5 cm if the particle is an electron.

Answer 1

Answer:

-1449.69404 N/C

Explanation:

u = Velocity of particle = $2.1\times 10^6\ m/s$

$\theta$ = Angle = 30°

x = Distance = 1.5 cm

m = Mass of electron = $9.11\times 10^{-31}\ kg$

q = Charge of electron = $-1.6\times 10^{-19}\ C$

In the case of projectile motion

$x=utcosA\\\Rightarrow t=\dfrac{x}{ucosA}$

The force of on the particle will balance the Electric force

$ma=qE\\\Rightarrow a=\dfrac{qE}{m}$

Now

$y=utsin\theta-\dfrac{1}{2}at^2\\\Rightarrow y=utsin\theta-\dfrac{1}{2}\dfrac{qE}{m}t^2$

If y = 0

$0=utsin\theta-\dfrac{1}{2}\dfrac{qE}{m}t^2\\\Rightarrow utsin\theta=\dfrac{1}{2}\dfrac{qE}{m}t^2\\\Rightarrow t=\dfrac{2musin\theta}{qE}$

$\dfrac{x}{ucosA}=\dfrac{2musin\theta}{qE}\\\Rightarrow E=\dfrac{2mu^2sin\theta cos\theta}{xq}\\\Rightarrow E=\dfrac{2\times 9.11\times 10^{-31}\times (2.1\times 10^6)^2\times sin30\times cos30}{1.5\times 10^{-2}\times (-1.6\times 10^{-19})}\\\Rightarrow E=-1449.69404\ N/C$

The electric field is -1449.69404 N/C

Answer 2

The required electric field is 1450 N/C

When the electron crosses the x-axis, its y-coordinate is zero. y = 0

If the electric field applied is E then the force acting on the electron:

F = eE

ma = eE ,

a = eE/m , here e is the charge on electron, m is the mass of electron.

If the field is applied towards positive y-axis then the force will be directed towards negative y-axis, since the elecrton is negatively charged.

If the electron leaves fromthe origin at a velocity u with an angle θ from x-axis then:

$u_x$ = ucosθ

$u_y$ = usinθ

Now from equation of motion:

$y=u_yt-\frac{1}{2}at^2$

y = utsinθ - 0.5eEt²/m      (i)

now, the time taken to reach x = 1.5 cm can be calculated by the horizontal component ucosθ:

t = x/ucosθ  substituting this in equation (i)

0= xtanθ  - 0.5eEx²/mu²cosθ²

E = mu²sinθcosθ/0.5ex  , here θ = 30

$E=\frac{ 9.11*10^{-31}*(2.1*10^6)^2*1/2*\sqrt{3}/2 }{0.5*1.6*10^{-19}*1.5*10^{-2}}$

E = 1450 N/C

The required electric field in the positive y-direction is 1450N/C

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