Question

How much work is required to bring a 10 μC charge from infinity to 5.00 meters away from a point charge of -2 μC?

Answer 1

The work done to move one charge from infinity to a given distance is:

`W=q(V_b-V_a)`

since

`V_b-V_a=kQ((1)/(r_b)-(1)/(r_a))`

and

`r_a=\infty\rightarrow(1)/(r_a)=0`

Then

`W=(kqQ)/(r_b)`

where:

`\begin{gathered} k=9\cdot10^9Nm^2C^(-2)\rightarrow eletric\text{ constant} \\ q=-2\mu C=-2\cdot10^(-6)C \\ Q=10\mu C=10\cdot10^(-6)C \\ r_b=5.00m \end{gathered}`

Plug all the data in the formula above:

`\begin{gathered} W=((9\cdot10^9)\cdot(-2\cdot10^(-6))\cdot(10\cdot10^(-6)))/((5.00)) \\ W=-36\cdot10^(-3) \\ W=-3.6\cdot10^(-4)J \end{gathered}`

So the final answer is:

`W=-3.6\cdot10^(-4)J`