Figure shows a narrow beam of electrons entering with a velocity of \(3\times10^7\ \text{m s}^{-1}\), symmetrically through the space between two parallel horizontal plates \(P_1P_1'\) and \(P_2P_2'\) kept \(2\ \text{cm}\) apart. If each plate is \(3\ \text{cm}\) long, calculate the potential difference \(V\) applied between the plates so that the beam just strikes the end \(P_2'\).
Show Hint
For an electron entering between parallel plates:
\[
E=\frac{V}{d},
\qquad
a=\frac{eV}{md}
\]
and the motion is projectile-like:
\[
x=u_xt,
\qquad
y=\frac12 at^2.
\]
Always calculate the time of flight first using horizontal motion and then use vertical motion to determine the required potential difference.
Concept:
When an electron enters the region between two oppositely charged parallel plates, it experiences a uniform electric field. The electric force acting on the electron produces a constant acceleration perpendicular to its initial direction of motion.
Thus, the motion of the electron is:
• Uniform motion along the horizontal direction.
• Uniformly accelerated motion along the vertical direction.
This is exactly analogous to the motion of a projectile.
The electric field between two parallel plates is
\[
E=\frac{V}{d},
\]
where
• \(V\) is the potential difference between the plates,
• \(d\) is the separation between the plates.
The force acting on the electron is
\[
F=eE,
\]
and hence the acceleration of the electron is
\[
a=\frac{eE}{m}
=\frac{eV}{md}.
\]
Step 1: Determine the time for which the electron remains between the plates.
The electron enters horizontally with velocity
\[
u_x=3\times10^7\ \text{m s}^{-1}.
\]
The length of each plate is
\[
l=3\ \text{cm}
=3\times10^{-2}\ \text{m}.
\]
Since there is no horizontal acceleration,
\[
t=\frac{l}{u_x}
\]
\[
t=
\frac{3\times10^{-2}}
{3\times10^7}
\]
\[
\boxed{
t=10^{-9}\ \text{s}
}
\]
Step 2: Calculate the vertical displacement of the electron.
The electron enters symmetrically between the plates.
Since the separation between the plates is
\[
d=2\ \text{cm}
=2\times10^{-2}\ \text{m},
\]
the electron has to travel a vertical distance of
\[
y=\frac{d}{2}
=1\times10^{-2}\ \text{m}
\]
to just strike the lower plate at the end \(P_2'\).
Initially, the vertical velocity is zero.
Therefore,
\[
y=\frac12 at^2.
\]
Substituting the values,
\[
10^{-2}
=
\frac12 a (10^{-9})^2.
\]
Hence,
\[
a
=
\frac{2\times10^{-2}}
{10^{-18}}
\]
\[
\boxed{
a=2\times10^{16}\ \text{m s}^{-2}
}
\]
Step 3: Use the expression for electric acceleration.
The acceleration of the electron is
\[
a=\frac{eV}{md}.
\]
Therefore,
\[
V=\frac{amd}{e}.
\]
Substituting
\[
a=2\times10^{16}\ \text{m s}^{-2},
\]
\[
m=9.1\times10^{-31}\ \text{kg},
\]
\[
d=2\times10^{-2}\ \text{m},
\]
\[
e=1.6\times10^{-19}\ \text{C},
\]
we get
\[
V=
\frac{(2\times10^{16})
(9.1\times10^{-31})
(2\times10^{-2})}
{1.6\times10^{-19}}.
\]
\[
V=
\frac{36.4\times10^{-17}}
{1.6\times10^{-19}}.
\]
\[
V=
22.75\times10^{2}.
\]
Therefore,
\[
\boxed{
V\approx2.28\times10^{3}\ \text{V}
}
\]
or
\[
\boxed{
V\approx2.3\times10^{3}\ \text{V}
}
\]
Hence, the potential difference that should be applied between the plates so that the electron beam just strikes the end \(P_2'\) is
\[
\boxed{
V\approx2.3\ \text{kV}
}
\]
