A charged particle leaves the origin with a speed of 2.94 x 106 m/s at an angle

Question

A charged particle leaves the origin with a speed of 2.94 x 106 m/s at an angle of 349 exists throughout the resglon above the x axis. A uniform electric field, given by E- al (a) Find Eo such that the particle will cross the x axis at x·1.37 ifthe partide is an electron. 0.001365 kN/C (b) Find Eo such that the particle will cross the x axis at x1.37 cm if the partide is a proton 251 eBook 22P A un orm line charge that has a linear charge density -2.8 nom is on the x als between x-0 to r·5.0 m. (a) What is its total charge? V nc Type here to search ac

Explanation / Answer

Given,

v0 = 2.94 x 10^6 m/s ; theta = 34 deg ; E = -E0 j

a)x = 1.37 cm = 0.0137 m

the x and y components of the intial velocity will be:

v0x = v0 cos(theta)

v0y = v0 sin(theta)

The displacement in the X direction would be:

X = v0x t

X = v0 cos(theta) t

t = X/v0 cos(theta)

Y = v0y t - 1/2 at^2

Y = v0 sin(theta) t -1/2 ay t^2

Net force, Fnet = m ay = q E

ay = q Ey/m

putting the value if ay in equation of Y we get

Y = v0 sin(theta)t - q Ey/2m t^2

Put the value of t, we get

Y = tan(theta) - qEy/2 m v0^2 cos^2(theta)

For Y = 0, solving for Ey we get

Ey = m v0^2 sin(2theta)/q x

Ey = 9.1 x 10^-31 x (2.94 x 10^6)^2 sin68/(1.6 x 10^-19 x 0.0137) = 3327.1 N/C

Hence, E = 3.327 kN/C

b)we just need to change the mass, rest all parameter will be the same.

Ey = 1.67 x 10^-27 x (2.94 x 10^6)^2 sin68/(1.6 x 10^-19 x 0.0137) = 6.106 x 10^6 N/C

Hence, Ey = 6.106 MN/C

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