Question

An $\alpha$ -particle of mass $6.4 \times 10^{-27} kg$ and charge $3.2 \times 10^{-19} C$ is situated in a uniform electric field of $1.6 \times 10^{5} Vm ^{-1} .$ The velocity of the particle at the end of $2 \times 10^{-2} m$ path when it starts from rest is

• A. $2 \sqrt{3} \times 10^{5} ms ^{-1}$
• B. $8 \times 10^{5} ms ^{-1}$
• C. $16 \times 10^{5} ms ^{-1}$
• D. $4 \sqrt{2} \times 10^{5} ms ^{-1}$

Answer 1

Answer: (D)

Force on a charge particle placed in a uniform electric field is given as,
$F=q E=m g$
$\Rightarrow E=\frac{m g}{q}$
From equation of motion,
$v^{2}=u^{2}+2 as=(0)^{2}+2 a s$
$\Rightarrow v=\sqrt{\frac{2 q E}{m}}$
$s=\sqrt{\frac{2 \times 3.2 \times 10^{-19} \times 1.6 \times 10^{5}}{6.4 \times 10^{-27}} \times 2 \times 10^{-2}}$
$\Rightarrow v =4 \sqrt{2} \times 10^{5} m / s$