Question

A particle of charge $$2 \mu C$$ and mass $$1.6 g$$ is moving with a velocity $$4 \hat{i} ms^{-1} $$. At $$t = 0$$ the particle enters in a region having an electric field $$\vec{E}$$ (in $$NC^{-1}) = 80 \hat{i} + 60 \hat{j}$$. Find the velocity of the particle at $$t = 5 s$$.

Answer 1

Given that :-

q = charge $$= 2 \mu C = 2 \times 10^{-6} C$$

mass $$= m = 1.6 g = 1.6 \times 10^{-3} kg$$

$$V = $$ velocity $$= 4 \hat{i} m/s$$

$$\vec{E} = (80 \hat{i} + 60 j ) N/C$$

Since particle has initial velocity only in x-direction

$$u_x = v = 4 m/s ; \, u_y = 0 m/s$$

Force $$= F = q \vec{E}$$

$$\Rightarrow \vec{F} = 2 \times 10^{-6} (80 \hat{i} + 60 j) = (160 \hat{i} + 120 j) \times 10^{-6}$$

$$\vec{a} = $$ acceleration $$= \vec{a} = \dfrac{\vec{F}}{m} = \dfrac{(160 \hat{i} + 120 j) \times 10^{-6}}{1.6 \times 10^{-3}}$$

$$\vec{a} = (100 \hat{i} + 75 \hat{j}) \times 10^{-3} = (a_x \hat{i} + a_y \hat{j})$$

time $$= t = 5 sec$$

$$v_x = u_x + a_x t = (4 + 100 \times 10^{-3} \times 5)$$

$$v_x = 4 + 0.5 = 4.5 m/s$$

$$v_y = u_y + a_y t = 0 + 75 \times 10^{-3} \times 5 = 0.375 m/s$$

Where $$u_x , v_x$$ are initial and final velocity in x-direction and $$u_y$$ and $$v_y$$ are initial and final velocity in y-direction.

$$\therefore $$ Final velocity $$= v = v_x \hat{i} + v_y \hat{j} = 4.5 \hat{i} + 0.375 j$$

$$v = (4.5 \hat{i} + 0.375 \hat{j}) m/s$$