# 8. Electric Charges by Integration

by M. Bourne

The force between charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.

So we can write:

`f(x)=(k\ q_1q_2)/x^2`

where *q*_{1} and *q*_{2} are in coulombs (C), *x* is in metres, the force is in newtons and *k* is a constant, *k* = 9 × 10^{9}.

It follows that the work done when electric charges move toward each other (or when they are separated) is given by:

`"Work"=int_a^b(k\ q_1q_2)/(x^2)dx`

### Example

An electron has a `1.6 × 10^-19\ "C"` negative charge. How much work is done in separating two electrons from `1.0\ "pm"` to `4.0\ "pm"`?

Answer

Recall: "`"pm"`" means **picometre**, or `10^-12` meters.

In this example,

`a = 1 ×10^-12\ "m"``b = 4 ×10^-12\ "m"`

`k = 9 × 10^9``q_1= q_2= 1.6 × 10^-19\ "C"`

So we have:

`"Work"=int_a^b (k\ q_1q_2)/(x^2) dx`

`=int_[1 times10^-12]^[4 times 10^-12] ((9 times 10^9)(-1.6 times 10^-19)^2)/(x^2) dx`

`=(2.304 times 10^-28)[-1/x]_[1 times 10^-12]^[4 times 10^-12]`

`=1.728 times 10^-16\ text[J]`

### Search IntMath

### Online Calculus Solver

This calculus solver can solve a wide range of math problems.

Go to: Online calculus solver