B Electric field on the equatorial line of a dipole

  • B
  • Thread starter Thread starter DrBanana
  • Start date Start date
  • Tags Tags
    Vector calculus
AI Thread Summary
The electric field on the line bisecting two equal and opposite charges is non-zero, while the potential along that line is zero. The relationship between electric field and potential is given by E = -dV/dr, but this only considers the radial component. A more comprehensive approach using vector calculus reveals that the electric field is a three-dimensional vector, requiring consideration of all components. The dipole potential in spherical coordinates is correctly expressed with V = (1/4πε₀)(p cos θ/r²), highlighting the importance of the correct power of r. In the equatorial plane, while the potential is zero, the derivative with respect to z remains non-zero, indicating a non-zero electric field.
DrBanana
Messages
54
Reaction score
4
##\vec{E}## on the line that perpendicularly bisects the segment that joins two equal and opposite charges is non-zero, as it should be. But the potential of any point along that line is zero. But we know know that ##E=-\frac{dV}{dr} ##, where V is approximately ##\frac{1}{4\pi \epsilon} \frac{pcos\theta}{r}## (if the charges are close together) where p is the magnitude of the dipole moment . If I differentiate that with respect to r and set ##\theta=\frac{\pi}{2}##, I still get E=0. What gives?
 
Physics news on Phys.org
DrBanana said:
What gives?
Nothing gives. The electric field is a three dimensional vector. You only calculated its radial component in the equatorial plane.
 
  • Like
Likes renormalize
DrBanana said:
##E=-\frac{dV}{dr} ##
That's wrong. It should be:$$\overrightarrow{E}=-\overrightarrow{\nabla}V=-\hat{r}\frac{\partial V}{\partial r}-\hat{\theta}\frac{1}{r}\frac{\partial V}{\partial\theta}-\hat{\phi}\frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}$$
 
renormalize said:
That's wrong. It should be:$$\overrightarrow{E}=-\overrightarrow{\nabla}V=-\hat{r}\frac{\partial V}{\partial r}-\hat{\theta}\frac{1}{r}\frac{\partial V}{\partial\theta}-\hat{\phi}\frac{1}{r\sin\theta}\frac{\partial V}{\partial\phi}$$
Ok I think I understand what happened. My physics book doesn't touch on vector calculus and only mentions ##E=-\frac{dV}{dr}##, however most of the forces in the book only have radial components anyway so it didn't matter. But that broke down here.
 
The dipole potential in spherical coordinates is $$V=\frac{1}{4\pi \epsilon_0}\frac{p~\cos\!\theta}{r^2}.$$Note the correct power of ##r## in the denominator. Also note that with ##z=r\cos\!\theta##, you have $$V=\frac{1}{4\pi \epsilon_0}\frac{p~z}{r^3}.$$This last expression can be considered to be the dipole potential in cylindrical coordinates. In the equatorial plane (##z=0##) the potential vanishes but not its derivative with respect to ##z## which you can easily calculate.
 
Thread 'Gauss' law seems to imply instantaneous electric field'
Imagine a charged sphere at the origin connected through an open switch to a vertical grounded wire. We wish to find an expression for the horizontal component of the electric field at a distance ##\mathbf{r}## from the sphere as it discharges. By using the Lorenz gauge condition: $$\nabla \cdot \mathbf{A} + \frac{1}{c^2}\frac{\partial \phi}{\partial t}=0\tag{1}$$ we find the following retarded solutions to the Maxwell equations If we assume that...
Hello! Let's say I have a cavity resonant at 10 GHz with a Q factor of 1000. Given the Lorentzian shape of the cavity, I can also drive the cavity at, say 100 MHz. Of course the response will be very very weak, but non-zero given that the Loretzian shape never really reaches zero. I am trying to understand how are the magnetic and electric field distributions of the field at 100 MHz relative to the ones at 10 GHz? In particular, if inside the cavity I have some structure, such as 2 plates...
Back
Top