Answer:
Work done, \(W=4.65\times 10^{-15}\ J\)
Explanation:
It is given that,
Magnitude of charge, \(q=3.1\times 10^{-16}\ C\)
Potential difference between two points, \(\Delta V=+15\ Volts\)
We have to find the magnitude of work done on that charge. The relationship between the work done and the potential difference is given by :
\(\Delta V=\dfrac{W}{q}\)
\(W=15\ V\times 3.1\times 10^{-16}\)
\(W=4.65\times 10^{-15}\ J\)
or
\(W=4.7\times 10^{-15}\ J\)
Hence, the magnitude of work done on the charge is \(4.65\times 10^{-15}\ J\)
