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 $\Delta V=-E x$, 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 $8.0 \times 10^{4} Vm ^{-1}$ directed along the positive $X$-axis. The proton undergoes a displacement of $0.50\, m$ 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 $0.50\, m$ starting from rest is

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

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 $v_{B}=\sqrt{\frac{2 \Delta K}{m}}$
$=\sqrt{\frac{2\left(6.4 \times 10^{-15} J \right)}{\left(1.66 \times 10^{-27} kg \right)}}$
$=2.77 \times 10^{6} ms ^{-1}$