Magnitude and direction of electric field at point P

[Ryomega]

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\frac{q}{r}

The Attempt at a Solution

1) V_tot = V_+q + V_-q

= kq(\frac{1}{x+a} - \frac{1}{x-a})

=\frac{2kqa}{x^2}

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

Thank you

 

[HallsofIvy]

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
\phi(x,y)= \frac{kq}{\sqrt{(x-a)^2+ y^2}}+ \frac{kq}{\sqrt{(x+a)^2a+ y^2}}
and the force vector the gradient of that:
\frac{\partial \phi}{\partial x}\vec{i}+ \frac{\partial \phi}{\partial y}\vec{j}

 

[Ryomega]

Ah, I see. Thank you