# Solving Charge Across Plates of a Capacitor w/ Negative Ions

• iVenky
In summary, the question is asking what would happen to the charge stored across parallel plates depleted of free electrons and containing negative ions. The student is confused about whether the charge would still be similar to placing a capacitor with a di-electric constant between them. They mention using the equation Q=CV, and then ask for help solving the question. The person helping them suggests using Gauss Law for charge on two surfaces, and provides a calculation for Q, the charge on the plates. The student then asks if there would still be some charge on the plates even if there is no supply or voltage.
iVenky

## 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?

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.

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)?

## 1. How do negative ions affect the charge across plates of a capacitor?

Negative ions play a crucial role in the charge distribution across the plates of a capacitor. They are attracted to the positive plate and repelled by the negative plate, causing a separation of charges and creating an electric field between the plates.

## 2. How does the distance between plates affect the charge across a capacitor with negative ions?

The distance between plates has a direct impact on the charge across a capacitor with negative ions. The closer the plates are, the stronger the electric field and the higher the charge density. As the distance increases, the electric field weakens and the charge density decreases.

## 3. Can negative ions be used to increase the capacitance of a capacitor?

Yes, negative ions can be used to increase the capacitance of a capacitor. This is because they increase the effective surface area of the plates, allowing for more charge storage. This is why some capacitors are designed with porous electrodes to attract more negative ions and increase the capacitance.

## 4. How do you calculate the charge across plates of a capacitor with negative ions?

The charge across plates of a capacitor with negative ions can be calculated using the formula Q = CV, where Q is the charge, C is the capacitance, and V is the potential difference between the plates. The capacitance can be calculated as C = εA/d, where ε is the permittivity of the material between the plates, A is the area of the plates, and d is the distance between the plates.

## 5. What is the role of dielectric materials in a capacitor with negative ions?

Dielectric materials are used in capacitors with negative ions to increase the capacitance by reducing the electric field between the plates. This allows for more charge to be stored on the plates. Dielectric materials also prevent the ions from reaching the plates and neutralizing the charge, thereby maintaining the separation of charges and the capacitor's ability to store energy.

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