1. Given data

• Mass of particle: $100\,\text{mg}$
• Electric field: $E = 1 \times 10^5\,\text{N C}^{-1}$
• Charge: $q = 40\,\mu\text{C}$
• Initial speed: $u = 200\,\text{m s}^{-1}$

The particle is thrown opposite to the electric field. Since the charge is positive, electric force acts along the electric field, i.e. opposite to the initial velocity. So the particle undergoes retardation.

2. Convert units

$m = 100\,\text{mg} = 100 \times 10^{-6}\,\text{kg} = 10^{-4}\,\text{kg}$

$q = 40\,\mu\text{C} = 40 \times 10^{-6}\,\text{C}$

3. Find electric force

$F = qE$

$F = (40 \times 10^{-6})(10^5) = 4\,\text{N}$

4. Find acceleration (retardation)

$a = \frac{F}{m} = \frac{4}{10^{-4}} = 4 \times 10^4\,\text{m s}^{-2}$

This acceleration is opposite to motion, so in kinematics we take

$a = -4 \times 10^4\,\text{m s}^{-2}$

5. Use equation of motion

At the turning point, final velocity becomes zero:

$v^2 = u^2 + 2as$

Putting $v=0$:

$0 = (200)^2 + 2(-4\times 10^4)s$

$0 = 40000 - 8\times 10^4 s$

$8\times 10^4 s = 4\times 10^4$

$s = \frac{4\times 10^4}{8\times 10^4} = 0.5\,\text{m}$

6. Check options

• A: $1\,\text{m}$ ❌
• B: $5\,\text{m}$ ❌
• C: $10\,\text{m}$ ❌
• D: $0.5\,\text{m}$ ✅

Hence, the particle travels $0.5\,\text{m}$ before coming to rest momentarily.