Derivation of the electric field from the potential

In summary, the conversation involves a student seeking help with a problem involving an electric dipole and the calculation of its electric potential and field. The student understands the concept but is struggling to remember or figure out how to derive the solution. The solution involves using the quotient rule to take the derivative and combining fractions. The final answer for the magnitude of the electric field is - 2*(Ke)*q * [ { (a^2) + (x^2) } / { ( (a^2)-(x^2) )^2 } ].
  • #1
stargirl22
3
0
:confused: 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 don't 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 } ]

Can Anyone please help? :D
 
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  • #2
Your potential should look something like this
[tex]q \left( \frac {1}{ {a-x}} - \frac {1}{ {a+x}}\right)[/tex]
so just combine the fractions.
 
  • #3
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 don't 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
stargirl22 said:
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 don't see how they derived the answer ...
Is your problem that you don't know how to take the derivative?
 
  • #5
You have to your the quotient rule!
 
  • #6
Sorry, I mean : You have to use the quotient rule!
 
  • #7
Quotient Rule

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

Attachments

  • Quotient Rule.JPG
    Quotient Rule.JPG
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FAQ: Derivation of the electric field from the potential

1. What is the relationship between electric field and potential?

The electric field is the measure of the force per unit charge acting on a charged particle at a given point in space. The potential, on the other hand, is a measure of the work required to move a charged particle from one point to another in an electric field. The relationship between the two is that the electric field is the negative gradient of the potential, meaning that the electric field points in the direction of decreasing potential.

2. How is the electric field derived from the potential?

The electric field can be derived from the potential using the formula E = -∇V, where E is the electric field, V is the potential, and ∇ is the gradient operator. This formula involves taking the partial derivative of the potential with respect to each of the spatial coordinates.

3. What is the significance of the electric field derived from the potential?

The electric field derived from the potential is significant because it allows us to understand and analyze the behavior of electrically charged particles in an electric field. It helps us determine the direction and strength of the forces acting on the particles, and thus their motion.

4. How does the electric field derived from the potential relate to the concept of electric potential energy?

The electric field and electric potential energy are closely related concepts. The electric field is a measure of the force acting on a charged particle, while electric potential energy is a measure of the potential energy a charged particle possesses by being in an electric field. The electric field derived from the potential can be used to calculate the change in electric potential energy as a charged particle moves through an electric field.

5. Are there any practical applications of deriving the electric field from the potential?

Yes, there are many practical applications of deriving the electric field from the potential. This concept is used in fields such as electrostatics, electromagnetism, and electronics. It is also used in the design and operation of various electrical devices, such as capacitors, electric motors, and generators.

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