The total work done of 0.018 joules is needed to move the charges apart and double the distance between them.
We have two electric charges q(A) = 1μc and q(B) = -2μc kept at a distance 0.5 meter apart.
We have to calculate much work is needed to move the charges apart and double the distance between them.
The potential energy of the system of two charges separated by a distance is given by -
[tex]U = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}[/tex]
In order to solve this question, it is important to remember the work - energy theorem which states -
"The change in the energy of the body is equal to work done on it"
Hence, using this work -energy theorem in the question given to us we get -
[tex]U_{f} -U_{i} =W_{net}[/tex]
In our case -
[tex]U_{f} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r}\\\\U_{i} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W=\frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r} - \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W = \frac{qQ}{4\pi\epsilon_{o}r} (\frac{1}{2} -1)\\\\W = 9\times 10^{9}\times \frac{1 \times 10^{-6} \times 2\times10^{-6} }{0.5} \times \frac{-1}{2}[/tex]
W = 0.018 joules
Hence, the total work done should be 0.018 joules.
To solve more question on potential energy, visit the link below -
https://brainly.com/question/15014856
#SPJ4