Question

A proton is released from rest in a uniform electric field of magnitude $8.0\times 10^{4}\, V/m$, directed along positive $x$-axis. The proton undergoes a displacement of $0.30\,m$ in the direction of the field. What is the change in electric potential of the proton as a result of this displacement?

• A. $2.4\times 10^{4}\,V$
• B. $4.8\times 10^{4}\,V$
• C. $-2.4\times 10^{4}\,V$
• D. $-1.2\times 10^{4}\,V$

Answer 1

Answer: (C)

$-2.4\times 10^{4}\,V$

Answer 2

Electric field of magnitude $8.0 \times 10^{4} V / m$
Displacement $=$ $0.30\, m$
Therefore, change in electric potential $=-E d$
$=-\left(8 \times 10^{4}\right)(0.30)$
$=-2.4 \times 10^{4} V$