Answer:
[tex]6.9\times 10^{-6} J[/tex]
Explanation:
We are given that
Charge=q=[tex]2.3\mu C=2.3\times 10^{-6} C[/tex]
[tex]1\mu C=10^{-6} C[/tex]
Potential difference=V=3 V
We know that
Work done=[tex]V\times q[/tex]
Using the formula
Work done by charge to move from the negative terminal to the positive terminal of battery=[tex]2.3\times 10^{-6}\times 3[/tex] J
Work done by charge to move from the negative terminal to the positive terminal of battery=[tex]6.9\times 10^{-6} J[/tex]
