A stream of a positively charged particles having $\frac{q}{ m }=2 \times 10^{11} \frac{ C }{ kg }$ and velocity $\vec{v}_0=3 \times 10^7 \hat{i} m / s$ is deflected by an electric field $18 kV / m$ The electric field exists in a region of $10 cm$ along $x$ direction Due to the electric field, the deflection of the charge particles in the $y$ direction is___ $mm$
1. Acceleration of the particles in the \(y\)-direction (\(a\)):
The force on the charged particles due to the electric field is:
\[
F = qE.
\]
Using \(F = ma\), the acceleration in the \(y\)-direction is:
\[
a = \frac{F}{m} = \frac{qE}{m}.
\]
Substitute the values:
\[
a = \left(2 \times 10^{11}\right) \cdot \left(1.8 \times 10^3\right) = 3.6 \times 10^{14} \, \text{m/s}^2.
\]
2. Time taken to cross the plates (\(t\)):
The time \(t\) to travel a distance of \(10 \, \text{cm} = 0.1 \, \text{m}\) along the \(x\)-direction is given by:
\[
t = \frac{d}{v_0},
\]
where \(d = 0.1 \, \text{m}\) and \(v_0 = 3 \times 10^7 \, \text{m/s}\).
Substitute the values:
\[
t = \frac{0.1}{3 \times 10^7} = \frac{1}{3 \times 10^8} \, \text{s}.
\]
3. Deflection in the \(y\)-direction (\(y\)):
The deflection in the \(y\)-direction is given by:
\[
y = \frac{1}{2} a t^2.
\]
Substitute the values:
\[
y = \frac{1}{2} \cdot \left(3.6 \times 10^{14}\right) \cdot \left(\frac{1}{3 \times 10^8}\right)^2.
\]
Simplify:
\[
y = \frac{1}{2} \cdot 3.6 \times 10^{14} \cdot \frac{1}{9 \times 10^{16}}.
\]
\[
y = \frac{3.6}{2} \cdot \frac{1}{9 \times 10^2} = \frac{1.8}{900} = 0.002 \, \text{m}.
\]
Convert to millimeters:
\[
y = 0.002 \, \text{m} = 2 \, \text{mm}.
\]
Thus, the deflection in the \(y\)-direction is:
\[
y = 2 \, \text{mm}.
\]
