# Show that if x a

Show that if x>>a....

## Homework Statement

I've got the following force generated by an electric field

$${F_x} = - KQq\frac{1}{{x\sqrt {{x^2} + {a^2}} }}$$

$${F_y} = \frac{{KQq}}{a}\left( {\frac{1}{x} - \frac{1}{{\sqrt {{x^2} + {a^2}} }}} \right)$$

## Homework Equations

I need to show that when x>>a:

$${F_x} = - \frac{{KQq}}{{{x^2}}}$$

$${F_y} = \frac{{KQqa}}{{2{x^3}}}$$

## The Attempt at a Solution

I think I'm on the right track but I'm stuck here:

$$\frac{1}{{\sqrt {{x^2} + {a^2}} }} = \frac{1}{x}{\left( {1 + \frac{{{a^2}}}{{{x^2}}}} \right)^{ - \frac{1}{2}}}$$

Have you heard of Taylor series? Why don't you try expanding the above term using taylor series and see which terms are negligible?

CompuChip
Homework Helper

You are indeed on the right track. When x >> a, then $\delta := \frac{a}{x}$ is very small. So you can expand
$$x \sqrt{1 + a^2 / x^2} = x \sqrt{1 + \delta^2}$$ around $\delta = 0$.

Thanks for your responses anirudh215 and CompuChip, but I really don't know what you mean by "expand". If you could show some steps I might have a clue.

ehild
Homework Helper

Do you know limits?

ehild

Thanks for your responses anirudh215 and CompuChip, but I really don't know what you mean by "expand". If you could show some steps I might have a clue.

Certain functions can be "expanded" around points, i.e. if you have a function f, and it is "expandable", then you can equate the value of that function about a point in terms of its derivatives.

The square root function above is one such function which is expandable. The Taylor series would be something like

$$\sqrt{1 + x} = 1 + \frac{x}{2} - \frac{x^{2}}{4.2!}$$....

Now, substitute $$\frac{a^{2}}{x^{2}}$$ as y, and expand similar to above. See what you can do from here. Note: the number of terms I have used in the expansion above is enough for you to complete the sum. Just look at what is and isn't negligible. I'm just doubtful whether you calculated Fy properly.