Answer to Question #173670 in Field Theory for Mya

Question

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.

Answer 1

Answer

Acceleration of electron

$a=\frac{eE}{m_e}=\frac{1.6\times10^{-19}(500)}{9.1\times10^{-31}}=88\times10^{12}m/s^2$

Time of motion

$t=\frac{2vsin30°}{a}=\frac{2(6.00\times10^6)sin30}{88\times10^{12}}=0.68\times10^{-7}s$

Maximum vertical distance

$s=\frac{(vsin30°)^2}{a}=\frac{(6\times10^{6}sin30°)^2}{88\times10^{12}}=0.1m$

Horizontal distance

$S_x=vcos30°(t) =6\times10^6(sin30°) (0.68\times10^{-7}=0.2m$