# Derivation of the electric field from the potential

1. Sep 26, 2004

### stargirl22

I am studying for a test and i can't figure out for the life of me how my book derived the solution for this problem I know it has to be basic i just dont see it...

An electric dipole consists of two charges of equal magnitude and opposite sign separated by a distance 2a... The dipole is along the x axis and is centered at the origin.
calculate V and Ex if point P is located anywhere between the two charges.

I understand the concept of this, and have calculated V, which is [ (2*(Ke)*q*x) / ((a^2) - (x^2)) ] and i know how to start the problem of Ex...

Ex = - (dV/dx) = - [ (2*(Ke)*q*x) / ( (a^2) - (x^2) ) ....

But I can't remember or figure out for the life of me how they got

= - 2*(Ke)*q * [ { (a^2) + (x^2) } / { ( (a^2)-(x^2) )^2 } ]

Last edited: Sep 26, 2004
2. Sep 26, 2004

### Tide

Your potential should look something like this
$$q \left( \frac {1}{ {a-x}} - \frac {1}{ {a+x}}\right)$$
so just combine the fractions.

3. Sep 26, 2004

### stargirl22

yes that is correct :D.. but i already got past that point and found the answer for the electric potential, V ... which was the sum of that equation...

and got (2KeqX) / ( a^2 - x^2) But that just gives me the potential... I needed to take it a step further and find the magnitude of the electric field from that... which is Ex ... which is = -dV/dx ... but i dont see how they derived the answer ...

= - 2*(Ke)*q * [ { (a^2) + (x^2) } / { ( (a^2)-(x^2) )^2 } ]

THANK YOU VERY MUCH THOUGH! :D

4. Sep 26, 2004

### Staff: Mentor

Is your problem that you don't know how to take the derivative?

5. Sep 28, 2004

### Simonnava

You have to your the quotient rule!!!!

6. Sep 28, 2004

### Simonnava

Sorry, I mean : You have to use the quotient rule!!!

7. Sep 28, 2004

### Simonnava

Quotient Rule

Please see the attached file. You will see how the quotient rule is require to get that answer.

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