Question

The electric potential at point X is 20 V and point Y is -40 V. Then, the work done by an external force is moving an electron from X to Y is:
A) $13 \times \mathop {10}\nolimits^{ - 8} J$
B) $9.6 \times \mathop {10}\nolimits^{ - 18} J$
C) $3.2 \times \mathop {10}\nolimits^{ - 8} J$
D) $30 \times \mathop {10}\nolimits^{ - 15} J$

Answer 1

Hint: In this question, the electric field is defined as the negative gradient of the electric potential so that the components of the electric field in any of the direction is the negative of the potential gradient is that direction. The external forces are forces which are caused by the external agents outside the system.

Complete step by step solution:
When we apply some external force on the system then the system will be experiencing some net force which will make the system displace and the work is being done on the system. Work done by an external force increases the energy of the system.
When external force is applied on the system then the system experiences some net force which makes the system displace and work is being done on the system. External force increases the energy of the system.
So, the test change in an electron,
Therefore, Work Done W = $q\Delta V$
$\Delta V = \mathop V\nolimits_B - \mathop V\nolimits_A $
$\mathop V\nolimits_B = - 40V$
$\mathop V\nolimits_A = - 20V$
$q = - 1.6 \times \mathop {10}\nolimits^{ - 19} C$
Now, work done W = $q\Delta V$
$ \Rightarrow W = \left( { - 1.6 \times \mathop {10}\nolimits^{ - 19} } \right)\left[ { - 40 - 20} \right]$
$ \Rightarrow W = 9.6 \times \mathop {10}\nolimits^{ - 18} J$
So, the work done by an external force is moving an electron from $X$ to $Y$ is $9.6 \times \mathop {10}\nolimits^{ - 18}J $

$\therefore$ The work done by an external force is moving an electron from X to Y is $9.6 \times \mathop {10}\nolimits^{ - 18}J $. Hence, option (B) is correct.

Note:
Many students may get confused about the sign formulas while answering these types of questions. The electric field is the negative differentiation of electric potential.
As the electric field is the differentiation of potential, we can express the potential as integration of the electric field. i.e.,
$v = - \smallint E.dx.$
When we find the electric field along $X$ we can also find the $Y$ and $Z$ axis. And we also differentiate $V$ (electric potential) with respect to the $Y$ and $Z$ axis.