Solving Charge Across Plates of a Capacitor w/ Negative Ions

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SUMMARY

The discussion focuses on the behavior of charge stored across the plates of a capacitor when negative ions are present between them. The key equation derived is Q^{(\text{plate})}=\frac{\Delta V\cdot \epsilon \cdot A}{l}-\frac{\rho_{\text{ions}}\cdot l \cdot A }{2}, which accounts for the electric field and the influence of negative ions. It is established that the presence of negative ions alters the charge distribution, leading to a different scenario compared to a standard capacitor with a dielectric constant. The analysis suggests that even with no external voltage supply, some charge remains on the plates due to the negative ions.

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  • Understanding of capacitor fundamentals, including charge storage and electric fields.
  • Familiarity with Gauss's Law for calculating electric fields and charge distributions.
  • Knowledge of dielectric materials and their effect on capacitance.
  • Basic algebra and manipulation of equations related to electric charge (Q=CV).
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iVenky
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Homework Statement


I have a material placed between parallel plates depleted of free electrons and contain negative ions. What would happen to the charge stored across the plates? Would it still be similar to placing a capacitor with a di-electric constant between them?

Homework Equations


Q=CV

The Attempt at a Solution


I am confused here because if I look at it in terms of E field, then the E field due to + charge on the plate and -ve ions cancel out at the other end of the plates. This means there would be unequal positive and negative charges on the parallel plates, which I am not sure makes sense.

upload_2019-1-27_22-28-8.png


Can you help me solve this question?
 

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iVenky said:
Would it still be similar to placing a capacitor with a di-electric constant between them?
It's a bit different.

iVenky said:
What would happen to the charge stored across the plates?
Try calculating yourself using two times Gauss Law for charge on two 'good' surfaces.
With my calculations, if I didn't make mistakes, I obtained:

$$Q^{(\text{plate})}=\frac{\Delta V\cdot \epsilon \cdot A}{l}-\frac{\rho_{\text{ions}}\cdot l \cdot A }{2}$$

with obvious notation.
 
Unconscious said:
It's a bit different.Try calculating yourself using two times Gauss Law for charge on two 'good' surfaces.
With my calculations, if I didn't make mistakes, I obtained:

$$Q^{(\text{plate})}=\frac{\Delta V\cdot \epsilon \cdot A}{l}-\frac{\rho_{\text{ions}}\cdot l \cdot A }{2}$$

with obvious notation.
If there is no supply (or voltage=0), you mean to say there would still be some charge on the plates (though the plates are neutral to begin with)?
 

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