Question

Two electric charges qa = 1. 0 μc and qb = - 2. 0 μc are located 0. 50 m apart. how much work is needed to move the charges apart and double the distance between them

Answer 1

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.

What s the formula to calculate the Potential Energy of a system of two charges (say 'q' and 'Q') separated by a distance 'r' ?

The potential energy of the system of two charges separated by a distance is given by -

$U = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}$

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 -

$U_{f} -U_{i} =W_{net}$

In our case -

$U_{f} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r}\\\\U_{i} = \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W=\frac{1}{4\pi\epsilon_{o} } \frac{qQ}{2r} - \frac{1}{4\pi\epsilon_{o} } \frac{qQ}{r}\\\\W = \frac{qQ}{4\pi\epsilon_{o}r} (\frac{1}{2} -1)\\\\W = 9\times 10^{9}\times \frac{1 \times 10^{-6} \times 2\times10^{-6} }{0.5} \times \frac{-1}{2}$

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 -

brainly.com/question/15014856

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