Question

An ion accelerated through a potential difference of 115 V experiences an increase in kinetic energy of 7.37 x 1017 J. Calculate the charge on the ion.

Answer 1

Answer: $6.408(10)^{-19} C$

Explanation:

This problem can be solved by the following equation:

$\Delta K=q V$

Where:

$\Delta K=7.37(10)^{-17} J$ is the change in kinetic energy

$V=115 V$ is the electric potential difference

$q$ is the electric charge

Finding $q$:

$q=\frac{\Delta K}{V}$

$q=\frac{7.37(10)^{-17} J}{115 V}$

Finally:

$q=6.408(10)^{-19} C$