At t.-0, a particle with q·0.8 C and m 0.05 kg is at x-yi· ·0 and is moving with

Question

At t.-0, a particle with q·0.8 C and m 0.05 kg is at x-yi· ·0 and is moving with v.-30 m/s in the +x direction. Ignore gravity; the only force on the particle comes from an electric field, E4.25 N/C in the direction of 40° above the +x axis in the x-y plane.tr0.25 s What is Fex? ex Submit Answer Tries 0/99 What is Fey? ey Submit Answer Tries 0/99 What is x? Submit Answer Tries 0/99 What is y? Submit Answer Tries 0/99 What is Vxr? Vyf = Submit Answer Tries 0/99 What is vy? Submit Answer Tries 0/99

Explanation / Answer

Given

charge q = 0.8 C, mass is m = 0.05 kg

initially at time ti =0 s , at xi=yi=zi=0

initial velocity along +s direction is vi = 30 m/s

electric field is E = 4.25 N/C , in the direction of 40 degrees above the x axis

later at time tf = 0.25 s

we know that F = Eq

and E = Ex i + Ey j = E cos theta i + E sin theta j = 4.25 cos40 i + 4.25 sin 40 j = 3.256 i + 2.732 j

so the force Fex = Ex*q = 3.256*0.8 N = 2.6048 N

Fey = Ey*q = 2.732*0.8 N = 2.1856 N

the accelertion of the charge particle along x,y directions is  

Fex = Ex*q = m*ax ==> ax = Ex*q/m = 2.6048/0.05 m/s2 = 52.096 m/s2

Fey = Ey*q = m*ay ==> ay = Ey*q/m = 2.1856/0.05 m/s2 = 43.712 m/s2

from equations of motion s

s = ut +0.5 at^2

Xf = ux*tf + 0.5*ax*tf^2

xf = 30*0.25+0.5*52.096*0.25^2 m = 9.128 m

and for yf  

yf = 0*0.25 + 0.5*43.712*0.25^2 m = 1.366 m

from equations of motions V = u+at

vxf = ux+ax*t = 30+52.096*0.25 m/s = 43.024 m/s

vyf = uy+ay*t = 0+43.712*0.25 m/s = 10.928 m/s

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