Question

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___ $mm$

Answer 1

Correct Answer: 2

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: \[ y = 2 \, \text{mm}. \]