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A material containing negative ions is placed between the plates of a charged capacitor

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1. 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?

2. Homework Equations
Q=CV

3. 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.

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Can you help me solve this question?
 

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Answers and Replies

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

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.
 
145
1
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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