Force on Point Charge (B) in Conductor Cavity

  • Thread starter Thread starter sspitz
  • Start date Start date
  • Tags Tags
    Conductor Force
sspitz
Messages
18
Reaction score
0
Suppose there is a cavity inside a conductor. Outside the conductor there is a point charge (A). E inside the cavity is zero because the field from the conductor and point charge cancel. That I believe.

Suppose I add a point charge (B) inside the cavity. Obviously, there is a radial field inside the cavity from the point charge (B). But won't the point charge (B) also mess up the distribution of charge on the surfaces of the conductor? Couldn't this new distribution produce a force on the point charge (B)?

My book treats as trivial that the force on B must be zero. I don't see it. Maybe an argument about the uniqueness of the potential function...
 
Physics news on Phys.org
Your book probably assumes the charge B is a "test" charge, small enough to not perturb the system. The idea is probably that the metal shell "shields" any outside fields.
 
Alas, no. It is definitely a charge of arbitrary Q for both A and B. I think in the specific example, A is at the center of a spherical cavity, but not centered in the conductor. Not sure if that matters and why or why not.
 
Have you figured it out? I'm curious. I think you can ignore the charge A, so that an arbitrary charge inside a metal cavity will not feel any electric field due to induced charge. Can you prove this is right?
 
Well, you definitely can't ignore B because it must affect the distribution of charge on the conductor. Maybe you can prove the new distribution still exerts no force on B.

I find it bizarre that ch. 3 of an intro EM book would have this problem and no explanation.

I have no rigorous proof that F on B is zero, but here is my best guess for B at the center of a spherical cavity.

Replace B with a very small conductor. The big conductor and B are equipotentials. One possible solution for the potential is the potential of a simple radial field from B to the surface of the cavity. Since the potential must be unique, this is the potential, and the field is radial in the limit as B becomes a point charge.

I don't even believe this argument, and I wrote it. I'm sure there is some very simple solution. I would appreciate someone pointing it out. Thanks.
 
##|\Psi|^2=\frac{1}{\sqrt{\pi b^2}}\exp(\frac{-(x-x_0)^2}{b^2}).## ##\braket{x}=\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dx\,x\exp(-\frac{(x-x_0)^2}{b^2}).## ##y=x-x_0 \quad x=y+x_0 \quad dy=dx.## The boundaries remain infinite, I believe. ##\frac{1}{\sqrt{\pi b^2}}\int_{-\infty}^{\infty}dy(y+x_0)\exp(\frac{-y^2}{b^2}).## ##\frac{2}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,y\exp(\frac{-y^2}{b^2})+\frac{2x_0}{\sqrt{\pi b^2}}\int_0^{\infty}dy\,\exp(-\frac{y^2}{b^2}).## I then resolved the two...
Hello everyone, I’m considering a point charge q that oscillates harmonically about the origin along the z-axis, e.g. $$z_{q}(t)= A\sin(wt)$$ In a strongly simplified / quasi-instantaneous approximation I ignore retardation and take the electric field at the position ##r=(x,y,z)## simply to be the “Coulomb field at the charge’s instantaneous position”: $$E(r,t)=\frac{q}{4\pi\varepsilon_{0}}\frac{r-r_{q}(t)}{||r-r_{q}(t)||^{3}}$$ with $$r_{q}(t)=(0,0,z_{q}(t))$$ (I’m aware this isn’t...
Hi, I had an exam and I completely messed up a problem. Especially one part which was necessary for the rest of the problem. Basically, I have a wormhole metric: $$(ds)^2 = -(dt)^2 + (dr)^2 + (r^2 + b^2)( (d\theta)^2 + sin^2 \theta (d\phi)^2 )$$ Where ##b=1## with an orbit only in the equatorial plane. We also know from the question that the orbit must satisfy this relationship: $$\varepsilon = \frac{1}{2} (\frac{dr}{d\tau})^2 + V_{eff}(r)$$ Ultimately, I was tasked to find the initial...
Back
Top