Question

Potential difference $(\Delta V)$ between two points $A$ and $B$ separated by a distance $x$, in a uniform electric field $E$ is given by $\Delta V=-Ex$, where $x$ is measured parallel to the field lines. If a charge $q_0$ moves from $A$ to $B$ the change in potential energy $(\Delta U)$ is given as $\Delta U=q_0\Delta V$. A proton is released from rest in uniform electric field of magnitude $8.0\times 10^{4}V{m}^{-1}$ directed along the positive $X$-axis. The proton undergoes a displacement of $0.50m$ in the direction of $E$.
Mass of a proton $=1.66\times 10^{-27}kg$
and charge on a proton $=1.6\times 10^{-19}C$

With the help of the passage given above, choose the most appropriate alternative for each of the following questions.
The velocity $\left(v_B\right)$ of the proton after it has moved $0.50m$ starting from rest is

• A. $1.6\times 10^{8}m{s}^{-1}$
• B. $2.77\times 10^{6}m{s}^{-1}$
• C. $2.77\times 10^{4}m{s}^{-1}$
• D. $1.6\times 10^{6}m{s}^{-1}$

Answer 1

Answer: (B)

As, $\Delta K=-\Delta U=6.4\times 10^{-15}J$
(from conservation of energy)
$\Delta K=\frac{1}{2}m{v}_B^2$
or $v_B=\sqrt{\frac{2\Delta K}{m}}$
$=\sqrt{\frac{2\left(6.4\times 10^{-15}J\right)}{\left(1.66\times 10^{-27}kg\right)}}$
$=2.77\times 10^{6}m{s}^{-1}$