Deriving an expression for the energy required to separate charges

[Taylor_1989]

Homework Statement

Hi guys, would just too make sure my derivation and insight to why is correct.
Question: a) only

Homework Equations

The Attempt at a Solution

$$dU=-wd_{ext}$$
$$dU=-F_{ext} \cdot dx$$
Now as the $F_{ext}$ is in the same direction and the direction vector { have not figured how to put direction vector in} then the equation becomes $dU=-Fdx$ this assume the direction is along the x-axis
So now if I intergrate both sides $\int_{U(a)}^{U\infty}dU=-\int_{a}^{\infty}Fdx$

Subbing in a factoring columbs force law I get:

$$U(\infty)-U(a)=-kq_1q_2[-\frac{1}{x}]_a^\infty$$

Now U infity is zero because there is no force acting on it anymore and so I am left with$-U(a)$ on the LHS on the RHS I am left with $-kq_1q_2\frac{1}{a}$
thus the two negative cancel and I am left with the electric poteinal energy. Is this correct?

 

[mfb]

If your answer has a positive sign, it is correct.Beyond the scope of your homework problem: (c) has a very interesting story. The naive way to improve the estimate leads to something that is not well-defined in mathematics, and you have to be careful how to do it properly to get the correct result.