Electric Dipole - Calulating the electic field from Potential

In summary, the electric field of a dipole is calculated using the Einstein notation for gradients and the Leibniz rule.
  • #1
smoking-frog
11
0

Homework Statement


Calculate the Electric field of a Dipole from its Potential.

[tex] \vec E=-\operatorname{grad}(\Phi_D)[/tex]

Homework Equations


[tex]\Phi_D(\vec R)=\frac{Q}{4\pi\epsilon_0} \cdot \frac{\vec d \vec R}{R^3}[/tex]

The Attempt at a Solution


Hi all!

I am trying to calculate the electric Field of a Dipole from its Potential.

Specifically, I am trying to understand the following equation:[tex]\operatorname{grad}(\Phi_D)=\frac{Q}{4\pi\epsilon_0R^3}\cdot((\vec d\cdot \vec R )\cdot \operatorname{grad}(\frac{1}{R^3})+\frac{1}{R^3} \cdot\operatorname{grad}(\vec d \cdot \vec R))[/tex]

I tried re-writing the equation according to the Einstein Notation:
[tex]\operatorname{grad}(\Phi_D)=\frac{Q}{4\pi\epsilon_0}\cdot\frac{\partial}{\partial R_j}\frac{d_k R_l}{(R_i R_i)^{\frac{3}{2}}}[/tex]

Is that correct? If so, how will I go about proving the above identity?

Thank you all for your help!
 
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  • #2
It is fairly straightforward to show, using the definition of gradient and partial derivatives, that gradient satisfies the Leibniz rule, [tex] \operatorname{grad}(fg) = f \operatorname{grad}(g) + g \operatorname{grad}(f) [/tex]
where f and g are scalars
 
  • #3
But they are not scalar, are they?

[tex] \vec d \text{ and } \vec R [/tex] are vectors...

The rule for vectors is, in my opinion, [tex]

\operatorname{grad}(\vec a \cdot \vec b)=(\vec a \cdot \nabla) \cdot \vec b+(\vec b \cdot \nabla) \cdot \vec a+\vec a \times (\nabla \times\vec b)+\vec b \times (\nabla \times\vec a)[/tex]
 
Last edited:
  • #4
smoking-frog said:
But they are not scalar, are they?

[tex] \vec d \text{ and } \vec R [/tex] are vectors...

Correct. But I thought in [itex] \phi [/itex] you had [itex] \vec{d} \cdot \vec{R} [/itex], which I believe is a scalar. Start from that and then note that [itex] \vec{d} [/itex]*is constant, so differentiating [itex] \vec{d} \cdot \vec{R} [/itex] is a bit simpler than that general formula suggests.
 
  • #5
Oh ok.

I actually tried that already, but the result was 0 for some reason...
But it's good to know I had the right idea and obviously just got lost in the math on the way.
I guess I'll simply try it again :)

Thanks for your help!
 

What is an electric dipole?

An electric dipole is a pair of equal and opposite charges that are separated by a distance. This separation of charges creates an electric field, which is a force that acts on other charges.

How do you calculate the electric field from potential?

The electric field can be calculated by taking the negative gradient of the electric potential. This means finding the partial derivatives of the potential with respect to each coordinate and multiplying by -1.

What is the formula for the electric field from potential?

The formula for the electric field from potential is E = -∇V, where E is the electric field, ∇ is the gradient operator, and V is the electric potential.

Can the electric field from potential be negative?

Yes, the electric field from potential can be negative. This indicates that the electric field is pointing in the opposite direction of the gradient of the potential.

What units are used to measure the electric field?

The electric field is measured in units of volts per meter (V/m) in the SI system. In other systems, it may be measured in units of newtons per coulomb (N/C) or pounds per volt (lb/V).

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