1. Given data

• Uniform electric field: $E = 10\,\text{N/C}$
• Electron enters midway between plates.
• Kinetic energy of electron: $K = 0.5\,\text{eV}$
• Length of plate: $L = 10\,\text{cm} = 0.1\,\text{m}$
• Charge of electron: $e = 1.6\times10^{-19}\,\text{C}$
• Mass of electron: $m = 9.1\times10^{-31}\,\text{kg}$

We assume the electron enters horizontally, so initially:

• horizontal velocity $= u$
• vertical velocity $= 0$

2. Find the initial horizontal speed from kinetic energy

Given $K = \frac12 m u^2 = 0.5\,\text{eV}$

Convert $0.5\,\text{eV}$ into joule: $K = 0.5\times 1.6\times10^{-19} = 0.8\times10^{-19}\,\text{J}$

So, $\frac12 m u^2 = 0.8\times10^{-19}$

$u^2 = \frac{2\times 0.8\times10^{-19}}{9.1\times10^{-31}}$

$u^2 \approx 1.76\times10^{11}$

$u \approx 4.2\times10^5\,\text{m/s}$

3. Find acceleration of electron inside the electric field

Force on electron: $F = eE$

So acceleration is $a = \frac{eE}{m}$

$a = \frac{1.6\times10^{-19}\times 10}{9.1\times10^{-31}}$

$a \approx 1.76\times10^{12}\,\text{m/s}^2$

4. Time spent by electron between the plates

Horizontal motion is uniform, so $t = \frac{L}{u} = \frac{0.1}{4.2\times10^5}$

$t \approx 2.38\times10^{-7}\,\text{s}$

5. Vertical velocity gained during motion in the field

Since initial vertical velocity is zero, $v_y = at$

$v_y = 1.76\times10^{12}\times 2.38\times10^{-7}$

$v_y \approx 4.2\times10^5\,\text{m/s}$

Thus, $v_y = u$

Also horizontal velocity remains unchanged: $v_x = u$

6. Angle of deviation at exit

The electron emerges with velocity components:

• $v_x = u$
• $v_y = u$

Hence, $\tan\theta = \frac{v_y}{v_x} = \frac{u}{u} = 1$

Therefore, $\theta = 45^\circ$

7. Final Answer

$\boxed{45}$