A positive charge particle of 100 mg is thrown in opposite direction to a uniform electric field of strength $$1 \times 10^5$$ N C$$^{-1}$$. If the charge on the particle is 40 $$\mu$$C and the initial velocity is 200 m s$$^{-1}$$, how much distance it will travel before coming to the rest momentarily

A positive charge particle of mass 100 mg is thrown opposite to a uniform electric field of $$1 \times 10^5$$ N C$$^{-1}$$. Charge = 40 $$\mu$$C, initial velocity = 200 m s$$^{-1}$$.

First, the particle is thrown opposite to the electric field. Since the particle is positively charged, the electric force acts along the field direction, opposing the motion.

$$F = qE = 40 \times 10^{-6} \times 1 \times 10^5 = 4 \text{ N}$$

$$a = \frac{F}{m} = \frac{4}{100 \times 10^{-6}} = \frac{4}{10^{-4}} = 4 \times 10^4 \text{ m s}^{-2}$$

Next, using $$v^2 = u^2 - 2as$$ (deceleration):

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

$$s = \frac{40000}{80000} = 0.5 \text{ m}$$

The correct answer is Option A: 0.5 m.