Question

An electron is projected with an initial speed $v_0=1.60\times {10}^{6}m{s}^{-1}$ into the uniform field between the parallel plates as shown in figure. Assume that the field between the plates is uniform and directed vertically downward, and that the field outside the plates is zero. The electron enters the field at a point midway between the plates. Mass of electron is $9.1\times {10}^{-31}kg$. For the condition that electron just misses the upper plate, the magnitude of electric field is :

• A. $124N{C}^{-1}$
• B. $364N{C}^{-1}$
• C. $224N{C}^{-1}$
• D. $520N{C}^{-1}$

Answer 1

The correct option is B $364N{C}^{-1}$

Passing between the charged plates, the electron feels a force upward
and just misses the top plate. The distance it travels in the
y-direction is 0.005 m.
in x direction electron is traveling 2m distance with velocity $v_0=1.60\times {10}^{6}ms$
Time of flight is $t=\frac{2m}{1.60\times {10}^{6}m{s}^{-1}}$
$=1.25\times {10}^{-8}s$
The initial y-velocity is zero.
Now, $y=v_{oy}t+\frac{1}{2}a{t}^2$
So, $0.005m=\frac{1}{2}a(1.25\times {10}^{-8}s{)}^2$
or $a=6.40\times {10}^{13}m{s}^{-2}$
But also $a=\frac{F}{m}=\frac{eE}{m_e}$
$E=\frac{(9.1\times {10}^{-32}kg)(6.40\times {10}^{13}m{s}^{-2})}{1.60\times {10}^{-19}C}=364N{C}^{-1}$