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 change in electric potential energy of the proton for displacement from $A$ to $B$ is

• A. $-6.4\times 10^{-19}J$
• B. $6.4\times 10^{-19}J$
• C. $-6.4\times 10^{-15}J$
• D. $6.4\times 10^{-15}J$

Answer 1

Answer: (C)

$\Delta U=q_0\Delta V=\left(1.6\times 10^{-19}C\right)\left(-4.0\times 10^{4}V\right)$
$=-6.4\times 10^{-15}J$