# Charge upon a parallel plate capacitor

• Pushoam
In summary, when a Gaussian cylindrical surface is taken between the two plates of a capacitor, the electric flux through it is zero due to the electric field being zero inside the material of the conductor. This leads to the conclusion that the two facing surfaces of the capacitor have equal and opposite charges. Additionally, the surface charge density on the inner surface is equal to that on the outer surface, resulting in the two plates of the capacitor having equal and opposite charges. For this to be true, the battery must provide unequal and opposite charges to the plates, with the outer surface charges also being opposite. The total charge supplied by the battery may not always be zero, as seen in the case of two capacitors in parallel with a net charge of 2

#### Pushoam

Homework Statement
Choose the correct options.
Each plate of a parallel plate capacitor has a charge Q on it. The capacitor is now connected to a battery. Now,
a) the facing surfaces of the capacitor have equal and opposite charges.
b) the two plates of the capacitor have equal and opposite charges.
c)the battery supplies equal and opposite charges to the two plates.
d) the outer surface of the plates have equal charges.
Relevant Equations
Gauss's law: $$\oint \vec E\cdot d\vec A= \frac{ q_{en}} {\epsilon_0}$$
a) if I take a Gaussian cylindrical surface whose circular area are present in the meat of the two plates of the capacitor, then the electric flux through this Gaussian surface is zero ( as the electric field inside the meatof the capacitor is zero and between the capacitors, electric field is perpendicular to the cylindrical surface).
So, charge enclosed by the Gaussian surface is zero.Hence the two facing surface of the capacitor have equal and opposite charges.

b) each plate of the conductor is equipotential. Hence the surface charge density on the inner surface is equal to that on the outer surface. Hence the two plates of the capacitor have equal and opposite charges.
c) to hold b) true the battery has to provide unqual and opposite charges to the two plates.
d) the outer surface charges of the plates have to be opposite.

Pushoam said:
Each plate of a parallel plate capacitor has a charge Q on it.
I take it that is intentional, that they don't just have the same magnitude of charge but also the same sign of charge.
Is that what you assumed?

haruspex said:
I take it that is intentional, that they don't just have the same magnitude of charge but also the same sign of charge.
Is that what you assumed?
Yes, initially both the plates have charge Q (same sign). Then, the capacitor is connected to the battery.

Pushoam said:
Yes, initially both the plates have charge Q (same sign). Then, the capacitor is connected to the battery.
Why don't your arguments for a) and b) apply before the plates are connected to the battery?
It might help to consider what a battery is similar to. Is it similar to a simple conductor? To a charged capacitor?

haruspex said:
Why don't your arguments for a) and b) apply before the plates are connected to the battery?
It might help to consider what a battery is similar to. Is it similar to a simple conductor? To a charged capacitor?
Before connecting to the battery, the plates are given charge Q as per question.
The battery is a device which maintains a given potential difference.

Is is true that the total charge supplied by a battery is always 0, i.e. if it gives Q to one plate, then it gives - Q to another plate?

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Pushoam said:
The battery is a device which maintains a given potential difference.
Yes.
Pushoam said:
the total charge supplied by a battery is always 0
That I'm less sure of. Supercapacitors are used as batteries. What would happen if, instead of a battery, a capacitor with charges +/-q were attached?

haruspex said:
What would happen if, instead of a battery, a capacitor with charges +/-q were attached?
The total charge supplied by the capacitor (as battery) is 0 as it gives charge q from the positive plate and -q from the negative plate.

Pushoam said:
The total charge supplied by the capacitor (as battery) is 0 as it gives charge q from the positive plate and -q from the negative plate.
So you'd have two capacitors in parallel, with one having a net zero charge but the other having a net charge of 2Q?

Taking the brown cylindrical surface as the Gaussian surface and applying the Gauss's law and considering the fact that electric field inside the material of conductor is 0 gives that surface charge densities on the facing surfaces are equal and opposite. Hence, the charges on the facing surfaces are q and -q.

Now, for outer surface, Electric field at P is calculated first using the Gaussian surface and applying Gauss's law and then, due to all the charged surfaces individually and then the two expressions are equated. $$\frac {q'}{A\epsilon_0} = \frac {q'}{2A\epsilon_0} +\frac {q}{2A\epsilon_0} +\frac {-q}{2A\epsilon_0} +\frac {q"}{2A\epsilon_0}$$ $$q' =q"$$

Hence, the inner surface charges of both plates are equal and opposite and the outer surface charges of both plates are equal and same.

The battery doesn't supply the charges. It creates the potential difference across the plates. As a result, electrons of plate at higher potential move towards plate at lower potential creating inner surface charge q and -q.

Now, the conservation of charge gives, $$q' +q -q+q' = 2Q \Rightarrow q' =Q$$

Hence, the correct options are a, c and d.

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Pushoam said:
View attachment 321947
Taking the brown cylindrical surface as the Gaussian surface and applying the Gauss's law and considering the fact that electric field inside the material of conductor is 0 gives that surface charge densities on the facing surfaces are equal and opposite. Hence, the charges on the facing surfaces are q and -q.

Now, for outer surface, Electric field at P is calculated first using the Gaussian surface and applying Gauss's law and then, due to all the charged surfaces individually and then the two expressions are equated. $$\frac {q'}{A\epsilon_0} = \frac {q'}{2A\epsilon_0} +\frac {q}{2A\epsilon_0} +\frac {-q}{2A\epsilon_0} +\frac {q"}{2A\epsilon_0}$$ $$q' =q"$$

Hence, the inner surface charges of both plates are equal and opposite and the outer surface charges of both plates are equal and same.

The battery doesn't supply the charges. It creates the potential difference across the plates. As a result, electrons of plate at higher potential move towards plate at lower potential creating inner surface charge q and -q.

Now, the conservation of charge gives, $$q' +q -q+q' = 2Q \Rightarrow q' =Q$$

Hence, the correct options are a, c and d.
I agree with your answers, but it is not clear that ##q'=Q##.
If we think of the battery as a charged capacitor, it would acquire some of the Q on each side, and now be carrying a net charge.

Pushoam
haruspex said:
I agree with your answers, but it is not clear that ##q'=Q##.
If we think of the battery as a charged capacitor, it would acquire some of the Q on each side, and now be carrying a net charge.
Let's consider the battery as a charged parallel plate capacitor. Then, the opposite charges are on the facing surfaces.
Now, if I draw a cylindrical Gaussian surface inside the capacitor such that its circular surfaces are in the material of the plates and cylindrical surface in the middle. Then, applying Gauss's law, one gets that the charges on the facing surfaces are opposite and equal, proving that the net charge is 0.

Pushoam said:
Let's consider the battery as a charged parallel plate capacitor. Then, the opposite charges are on the facing surfaces.
Now, if I draw a cylindrical Gaussian surface inside the capacitor such that its circular surfaces are in the material of the plates and cylindrical surface in the middle. Then, applying Gauss's law, one gets that the charges on the facing surfaces are opposite and equal, proving that the net charge is 0.
If I connect a capacitor C1 with Q on each plate in parallel with uncharged capacitor C2, how will the charge redistribute? You seem to be claiming it won’t.

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Pushoam said:
View attachment 321947Now, the conservation of charge gives, $$q' +q -q+q' = 2Q \Rightarrow q' =Q$$

Hence, the correct options are a, c and d.
For a capacitor connected to a battery the conservation of charge on the capacitor is not granted and not required.
You showed that the charges on the inner faces are opposite and equal. Also that q'=q''.
But the fact that these last two are equal to Q does not follow. I am not saying it cannot be true, but you need something else to prove it, if true.

Pushoam
haruspex said:
If I connect a capacitor C1 with Q on each plate in parallel with uncharged capacitor C2, how will the charge redistribute? You seem to be claiming it won’t.
haruspex said:
If we think of the battery as a charged capacitor, it would acquire some of the Q on each side, and now be carrying a net charge.

Let's say that the battery capacitor has capacitance Cb, initial charge Qb, and potential difference Vb and the capacitor has charge Q on each plate.
After the connection and coming to equilibrium, the system has the charge configuration given in the figure and potential difference V'b.
Calculations similar to those in post #9 show that outer surface charges on the plates of battery are same.

Now, $$q_b = C_b V'_b ~~\text{and}~~q = CV'_b$$ $$\Rightarrow q = \frac {q_b C}{C_b} ~~\text{and} ~~q_b = \frac{Q_B}{1+\frac C {C_B}}$$

So, after the connection, potential difference and facing surface charges of each plate changes.

The charge Q, too, redistributes on the outer surface of all plates as the connecting plates of the battery and capacitor are equipotential. Could you help me in finding the distribution?

On the other hand, an ideal battery is supposed to maintain the potential difference. Here, the potential difference changes. So, the capacitor battery is not an ideal battery.

Pushoam said:
On the other hand, an ideal battery is supposed to maintain the potential difference
If Q' out of the initial Q on each plate moves into the battery, the potential difference between the plates is unaffected by that.

nasu and Pushoam
haruspex said:
If Q' out of the initial Q on each plate moves into the battery, the potential difference between the plates is unaffected by that.
This I have understood.
The thing is I am not able to prove that this Q' is nonzero. I just know that the charge will distribute bcause these plates are conducting.

Pushoam said:
This I have understood.
The thing is I am not able to prove that this Q' is nonzero. I just know that the charge will distribute bcause these plates are conducting.
It will be nonzero if the battery acts at least a tiny bit like a capacitor.

haruspex said:
It will be nonzero if the battery acts at least a tiny bit like a capacitor.
How can this be shown mathematically?

Pushoam said:
How can this be shown mathematically?
By solving the question I posed in post #12.

Pushoam

## 1. What is the formula for calculating the charge on a parallel plate capacitor?

The formula for calculating the charge on a parallel plate capacitor is Q = CV, where Q is the charge in Coulombs, C is the capacitance in Farads, and V is the voltage across the capacitor in Volts.

## 2. How does the distance between the plates affect the charge on a parallel plate capacitor?

The distance between the plates has an inverse relationship with the charge on a parallel plate capacitor. As the distance between the plates increases, the charge decreases, and vice versa.

## 3. What is the role of the dielectric material in a parallel plate capacitor?

The dielectric material, which is placed between the plates of a parallel plate capacitor, acts as an insulator and helps to increase the capacitance of the capacitor. It does this by reducing the electric field between the plates.

## 4. How does the area of the plates affect the charge on a parallel plate capacitor?

The area of the plates has a direct relationship with the charge on a parallel plate capacitor. As the area of the plates increases, the charge also increases, and vice versa.

## 5. What happens to the charge on a parallel plate capacitor when the voltage is increased?

When the voltage across a parallel plate capacitor is increased, the charge on the capacitor also increases. This is because the voltage is directly proportional to the charge, according to the Q = CV formula.

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