Question

An electron is projected with velocity 10 m/s at an angle 0 (= 30°) with horizontal in a region of uniform electric field of 5000 N/C vertically upwards. The maximum distance covered by an electron in vertical direction above its initial level is

Answer 1

Given $E=5000 \;N/C$ and $v_0=10\;m/s,\theta =30^o$. Tghe force on the electron acts vertically downwards.

When the vertical velocity of the electron becomes 0,

$0^2=v_0^2\sin^2 \theta -2ah$ or

$h=\frac{v_0^2\sin^2 \theta}{2a}$

The deceleration of the electron is $a=\frac{Eq}{m} \\ a=\frac{5000(1.602*10^{-19})}{9.11*10^{-31}} \\ a=8.79*10^{14}$

The maximum distance above the initial level is

$h=\frac{10^2(0.5)^2}{2*8.79*10^{14}}\\ h=1.42*10^{-16} \;m$

Answer 2

Explanation:

all of the kinetic energy in the vertical direction converts to potential energy jn that direction at that point distance covered by the particle will be maximum

hence

1/2m(vsin30°)²=Eeh

where m=mass of electron

e=charge of electron

h=maximum distance in the vertical direction

as we know E=500N/C and all the other values substituting them in the above equation we get

h=14.2cm