Q.
An electron is released from the bottom plate $A$ as shown in the figure $(E = 10^{4}N/C).$ The velocity of the electron when it reaches plate $B$ will be nearly equal to
A) $0.85 \times 10^{7}\ m/s$
B) $1.0 \times 10^{7}\ m/s$
C) $1.25 \times 10^{7}\ m/s$
D) $1.65 \times 10^{7}\ m/s$
Explanation
(a) The force acting on the electron $= e.E$
Acceleration of the electron $= \frac{eE}{m}$
[image:1_image290.emf]
$u = 0,v = ?\ \ v^{2} - u^{2} = 2aS$
$S = 2 \times 10^{- 2}m$
$\therefore v^{2} = 2\left( \frac{e}{m} \right)E \times 2 \times 10^{- 2}m$
$\frac{e}{m} = 1.76 \times 10^{11}\ coulomb/kg$
$\therefore v^{2} = 2 \times 1.76 \times 10^{11} \times 10^{4} \times 2 \times 10^{- 2}$
$= 7.04 \times 10^{13} = 70.4 \times 10^{12}$
$\therefore v \approx 0.85 \times 10^{7}\ m/s$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.