Magnitude and direction of electric field at point P

  • Thread starter Ryomega
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  • #1
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Homework Statement



An electric dipole is defined as two point charges separated by distance L.
Consider point P with coordinates (x,0).
A dipole along the x axis is separated by distance a from the origin. q- is closer to point P
Find:

1) Electrostatic potential due to dipole at point P
2) Magnitude and direction of electric field E at point P in terms of dipole moment and distance x

Homework Equations



V = k[itex]\frac{q}{r}[/itex]

The Attempt at a Solution



1) Vtot = V+q + V-q

= kq([itex]\frac{1}{x+a}[/itex] - [itex]\frac{1}{x-a}[/itex])

=[itex]\frac{2kqa}{x^2}[/itex]

2) I know that taking the divergence of [itex]\frac{2kqa}{x^2}[/itex] will give a scalar. Is this divergence going to be [itex]\frac{d}{dx}[/itex]? How do I go about getting the magnitude and direction from there?

Thank you
 

Answers and Replies

  • #2
HallsofIvy
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The value of the gradient is the magnitude of the vector. Since both poles are on the x-axis and your given point is (x, 0), the vector will point along the x-axis to the nearest pole. That means that for x> 0 it points toward (a, 0) and for x< 0, it points toward (-a, 0).

More generally, if the point could be any point in the xy-plane, (x, y), rather than just (x, 0), the potential function would be
[tex]\phi(x,y)= \frac{kq}{\sqrt{(x-a)^2+ y^2}}+ \frac{kq}{\sqrt{(x+a)^2a+ y^2}}[/tex]
and the force vector the gradient of that:
[tex]\frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}[/tex]
 
  • #3
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Ah, I see. Thank you
 

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