Electromagnetism, force between dipole and grounded plane

AI Thread Summary
The discussion revolves around calculating the force exerted by an electric dipole on a grounded plane. It clarifies that the grounded plane, despite having no net charge, can still interact with the dipole due to the concept of image charges. The dipole is treated as two point charges, which induce an equal and opposite charge in the grounded plane. The potential field and interaction energy can be expressed using the dipole moments and their separation, leading to a derived force equation. The final expression for the radial force takes into account the distance and angle factors, emphasizing the importance of classical electromagnetism principles.
Antti
Messages
26
Reaction score
0

Homework Statement



An electric dipole is located at a certain distance from a grounded plane. What force does the dipole exert on the plane?

(The answer is to be expressed as an equation. No data were given, only the above text)

Homework Equations



I am not sure about this, but the course is about "classical" electromagnetism. I would suspect that coulombs law should be used since the question is about the force between charged particles/objects.

F = \frac{Q_{1}Q_{2}}{4 \pi \epsilon_{0} r^{2}}

The Attempt at a Solution



At first I just thought that the grounded plane would have no net charge and thus the dipole and plane could not affect each other. I now know that this is isn't true but I'm not sure why. A coursemate told me that the dipole should be treated as two point charges and that they would have equal but opposite charges (mirrored) in the grounded plane. Unfortunately I didn't get the chance to ask him further questions.

Thankful for help
 
Last edited:
Physics news on Phys.org
If M1=Qd is the dipole moment (Q :charge, d:separation as a vector) the potential field is
V(r)=1/(4*pi*e) M1.grad(1/r)
Having a ground plane, this can be eliminated if you consider an image dipole (symmetrically placed under the ground plane). of moment M2=-M1.
Then the interaction energy will be of the order
W=M1*M2/(4 pi eps r^3)*(angle factor)
where r is the distance between the two dipoles.
The radial force will be
F=-dW/dr=3m1m2/(4 pi eps r^4)*(angle factor)
For the angle factor look at classic SMYTHE-Static&DynamicElectricity- McGraw 1968, p.7
 
I multiplied the values first without the error limit. Got 19.38. rounded it off to 2 significant figures since the given data has 2 significant figures. So = 19. For error I used the above formula. It comes out about 1.48. Now my question is. Should I write the answer as 19±1.5 (rounding 1.48 to 2 significant figures) OR should I write it as 19±1. So in short, should the error have same number of significant figures as the mean value or should it have the same number of decimal places as...
Thread 'A cylinder connected to a hanging mass'
Let's declare that for the cylinder, mass = M = 10 kg Radius = R = 4 m For the wall and the floor, Friction coeff = ##\mu## = 0.5 For the hanging mass, mass = m = 11 kg First, we divide the force according to their respective plane (x and y thing, correct me if I'm wrong) and according to which, cylinder or the hanging mass, they're working on. Force on the hanging mass $$mg - T = ma$$ Force(Cylinder) on y $$N_f + f_w - Mg = 0$$ Force(Cylinder) on x $$T + f_f - N_w = Ma$$ There's also...
Back
Top