How does an electret retain its charge for years without decreasing?

  • Thread starter Thread starter arul_k
  • Start date Start date
AI Thread Summary
Electrets can retain their charge for years due to their permanent electric polarization, which remains intact even after discharge. Discharge occurs when air ions neutralize the electret's electric field, but this does not affect the underlying polarization. Recharging happens by removing these air ions, restoring the electret's electric field. Shorting the electret does not diminish its polarization charge, as the induced polarization from the shorting wire dissipates once removed. Overall, electrets can maintain their charge indefinitely, especially when kept in a vacuum.
arul_k
Messages
95
Reaction score
1
How is an electret capable of recharging itself after discharge. I came across an article that stated tha electrets are capable of retaining their charge for years and any number of discharges does not decrease their charge. how does this happen?
 
Physics news on Phys.org
arul_k said:
How is an electret capable of recharging itself after discharge. I came across an article that stated tha electrets are capable of retaining their charge for years and any number of discharges does not decrease their charge. how does this happen?

In the case of a bar electret (e.g. one made out of barium titanate) "discharge" might occur when ions in the air are attracted to the ends, thus neutralizing the electret's electric field. This process does not change the bar's permanent, electric polarization. When the electret is "recharged", by stripping away the air ions, the bar's polarization field is restored until "discharge" occurs again. Kept in a vacuum, the electret should remain "charged" indefinitely.
 
GRDixon said:
In the case of a bar electret (e.g. one made out of barium titanate) "discharge" might occur when ions in the air are attracted to the ends, thus neutralizing the electret's electric field. This process does not change the bar's permanent, electric polarization. When the electret is "recharged", by stripping away the air ions, the bar's polarization field is restored until "discharge" occurs again. Kept in a vacuum, the electret should remain "charged" indefinitely.

Thanks for your reply. I was thinking in terms of a spark discharge that would occour between the +ve and -ve terminals of the electret, when shorted. The process you suggest seems more like a neutralisation of the electric field.
 
arul_k said:
Thanks for your reply. I was thinking in terms of a spark discharge that would occour between the +ve and -ve terminals of the electret, when shorted. The process you suggest seems more like a neutralisation of the electric field.

The key to the puzzle may be that the charges responsible for an electret's field are POLARIZATION charges, and are therefore bound charges. As such, shorting the poles as suggested will not result in a loss of the polarization charge. The shorting wire, which does contain conduction electrons, no doubt becomes electrically polarized while in contact (or in close proximity) with the electret's poles. But once removed, it loses this induced polarization and the electret's intrinsic electric field returns to its nominal value. Any conduction charges that "rub off" of the shorting wire, and stick to the electret's positive pole, might be relatively unimportant, as the electret is a dielectric.

But all of the above should be taken with a grain of salt, as I am not experienced with these fascinating devices.
 
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...
Maxwell’s equations imply the following wave equation for the electric field $$\nabla^2\mathbf{E}-\frac{1}{c^2}\frac{\partial^2\mathbf{E}}{\partial t^2} = \frac{1}{\varepsilon_0}\nabla\rho+\mu_0\frac{\partial\mathbf J}{\partial t}.\tag{1}$$ I wonder if eqn.##(1)## can be split into the following transverse part $$\nabla^2\mathbf{E}_T-\frac{1}{c^2}\frac{\partial^2\mathbf{E}_T}{\partial t^2} = \mu_0\frac{\partial\mathbf{J}_T}{\partial t}\tag{2}$$ and longitudinal part...
Thread 'Recovering Hamilton's Equations from Poisson brackets'
The issue : Let me start by copying and pasting the relevant passage from the text, thanks to modern day methods of computing. The trouble is, in equation (4.79), it completely ignores the partial derivative of ##q_i## with respect to time, i.e. it puts ##\partial q_i/\partial t=0##. But ##q_i## is a dynamical variable of ##t##, or ##q_i(t)##. In the derivation of Hamilton's equations from the Hamiltonian, viz. ##H = p_i \dot q_i-L##, nowhere did we assume that ##\partial q_i/\partial...
Back
Top