A stream of a positively charged particles having $$\frac{q}{m} = 2 \times 10^{11} \text{ C kg}^{-1}$$ and velocity $$\vec{v}_0 = 3 \times 10^7 \hat{i} \text{ m s}^{-1}$$ is deflected by an electric field $$1.8\hat{j} \text{ kV m}^{-1}$$. 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.

A stream of positively charged particles with $$\frac{q}{m} = 2 \times 10^{11} \text{ C/kg}$$ enters with velocity $$\vec{v}_0 = 3 \times 10^7 \hat{i} \text{ m/s}$$ into a region of electric field $$\vec{E} = 1.8 \hat{j} \text{ kV/m} = 1800 \hat{j} \text{ V/m}$$, extending 10 cm along the $$x$$-direction.

Find the time spent in the electric field.

$$t = \frac{d}{v_0} = \frac{0.1}{3 \times 10^7} = \frac{1}{3 \times 10^8} \text{ s}$$

Find the acceleration in the $$y$$-direction.

$$a = \frac{qE}{m} = \frac{q}{m} \times E = 2 \times 10^{11} \times 1800 = 3.6 \times 10^{14} \text{ m/s}^2$$

Find the deflection in the $$y$$-direction.

$$y = \frac{1}{2}at^2 = \frac{1}{2} \times 3.6 \times 10^{14} \times \left(\frac{1}{3 \times 10^8}\right)^2$$

$$= \frac{1}{2} \times 3.6 \times 10^{14} \times \frac{1}{9 \times 10^{16}} = \frac{3.6}{18} \times 10^{-2} = 0.2 \times 10^{-2} = 2 \times 10^{-3} \text{ m}$$

$$y = 2 \text{ mm}$$

The answer is $$\boxed{2}$$ mm.