Question #173670

An electron is projected into a uniform electric field that has a magnitude of 500 N/C.

The direction of the electric field is vertically upward. The initial velocity of the

electron has a magnitude of 6.00 × 10^6 m/s, and its direction is at an angle of 30°

above the horizontal. Find:

I. The maximum distance the electron rises vertically above its initial elevation.

II. After what horizontal distance does the electron return to its original elevation.


1
Expert's answer
2021-03-23T08:08:29-0400

Answer

Acceleration of electron

a=eEme=1.6×1019(500)9.1×1031=88×1012m/s2a=\frac{eE}{m_e}=\frac{1.6\times10^{-19}(500)}{9.1\times10^{-31}}=88\times10^{12}m/s^2


Time of motion

t=2vsin30°a=2(6.00×106)sin3088×1012=0.68×107st=\frac{2vsin30°}{a}=\frac{2(6.00\times10^6)sin30}{88\times10^{12}}=0.68\times10^{-7}s


Maximum vertical distance

s=(vsin30°)2a=(6×106sin30°)288×1012=0.1ms=\frac{(vsin30°)^2}{a}=\frac{(6\times10^{6}sin30°)^2}{88\times10^{12}}=0.1m


Horizontal distance

Sx=vcos30°(t)=6×106(sin30°)(0.68×107=0.2mS_x=vcos30°(t) =6\times10^6(sin30°) (0.68\times10^{-7}=0.2m





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