Capacitor Formation with Unequal Charges: A Potential Difference Dilemma

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In summary, the potential difference across two plates is proportional to the product of the charges on the plates. The capacitance of a parallel-plate capacitor increases as the distance between the plates decreases.
  • #1
gracy
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Homework Statement


Two plates (area=s) charged to +q1 and +q2 (q2<q1)are brought closer to form a capacitor of capacitance C.The potential difference across the plates is

2. Homework Equations

##V##=##\frac{Q}{C}##

The Attempt at a Solution


[/B]
I don't know how are these plates going to form a capacitor as both of them have same type of (i.e positive)charge and in order to form a capacitor two conducting plates with equal and opposite charges are required.
 
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  • #2
What have you learned about a parallel-plate capacitor? What is the capacitance if the area of the plates is s and the distance between them is d? Does the capacitance depend on the charge of the plates?
 
  • #3
I would be inclined to look at the net electric field between the plates.
 
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  • #4
gneill said:
I would be inclined to look at the net electric field between the plates.
##\vec{E}##=##\frac{σ}{ε0}##=##\frac{Q}{ε0A}##
 
  • #5
What if one would to consider two capacitors one charged with a q1 charge and the other with a q2 charge?
 
  • #6
gracy said:
##\vec{E}##=##\frac{σ}{ε0}##=##\frac{Q}{ε0A}##
That's the idea, but check your formula for the electric field... a sheet of charge has two sides so you need to take that into account when you apply Gauss' Law to find the field.

edit: See for example: Hyperphysics: Electric Feild, Flat Sheets of Charge

Now, you have two sheets of charge with different charge densities...
 
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  • #7
andrevdh said:
two capacitors
Two capacitors or two plates of a capacitor?
 
  • #8
gracy said:

Homework Statement


Two plates (area=s) charged to +q1 and +q2 (q2<q1)are brought closer to form a capacitor of capacitance C
...
in order to form a capacitor two capacitors conducting plates with equal and opposite charges are required.
ehild said:
Does the capacitance depend on the charge of the plates

Let q1 = 1, q2 = 2 for simplicity.

If the plates are far apart (far enough that they don't influence each other significantly), can you imagine the charges are evenly distributed over the surfaces ?

What happens if they are brought closer together ?
 
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  • #9
BvU said:
What happens if they are brought closer together ?
Capacitance increases.
I don't want to sound skeptical that's why I avoided question mark!It does not mean I am confident about it.
 
  • #10
My thought process is
For a parallel plate capacitor
##C##=##\frac{Aε0}{d}##
As d (distance between the two plates )decreases,C i.e Capacitance should increase.
 
  • #11
Am I going in right/correct direction?
 
  • #12
gracy said:
Am I going in right/correct direction?
I still feel that you'd be better to look at the net field between the plates. Gauss' Law will give you the answer.
 
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  • #13
gneill said:
check your formula for the electric field..
##E##=##\frac{σ}{2ε0}##
Right?
 
  • #14
gracy said:
##E##=##\frac{σ}{2ε0}##
Right?
That will give you the field between the plates due to one plate with net charge density σ. You have two plates with different σ's. Draw a sketch and pay attention to the field directions. How do they sum?
 
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  • #15
ETF.png


Electric field due to one plate having charge density σ1 in between the two plates(we will avoid electric field at other places)
##E_1##=##\frac{σ1}{2ε0}##

##E_2##=##\frac{σ2}{2ε0}##
gneill said:
How do they sum?
Vector addition of ##E_1## and ##E_2##
 
  • #16
Okay, so assuming that you choose an appropriate direction for the net field, what's an expression for its magnitude?
 
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  • #17
gneill said:
what's an expression for its magnitude?
##\frac{σ1}{2ε0}##-##\frac{σ2}{2ε0}##
as Q1>Q2
Hence σ1>σ2
 
  • #18
Good. But you can expand that a bit using the problem's given variables: s, q1, q2. Usually you want to express any results using the given parameters. You can pretty it up by factoring out the common variables.

Next, given a constant electric field, how do you find the potential difference between two locations?
 
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  • #19
##\frac{Q1}{2ε0s}##-##\frac{Q2}{2ε0s}##
 
  • #20
Keep going...
 
  • #21
gneill said:
Next, given a constant electric field, how do you find the potential difference between two locations?
But it is not mentioned that electric field is constant,is it?
 
  • #22
gracy said:
But it is not mentioned that electric field is constant,is it?
The field from a relatively large sheet of charge is uniform. Here "relatively large" means that for the location of interest the distance from a face of the sheet is small compared to the size of the sheet, and it's not too close to the edges of the sheet where "edge effects" interfere with the uniformity.
 
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  • #23
So I should consider electric field to be uniform in this case?
 
  • #24
gracy said:
So I should consider electric field to be uniform in this case?
Yes, clearly. There are two parallel sheets of charge on plates forming a capacitor.
 
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  • #25
gneill said:
how do you find the potential difference between two locations?
Electric field multiplied by r?
 
  • #26
Answer given in my book is
##\frac{q1-q2}{2C}##
 
  • #27
gracy said:
Electric field multiplied by r?
Sure. You make it look like a guess though. It should be an expression that's given in your text and notes (although they're more likely to use "d" rather than "r").

gracy said:
Answer given in my book is
##\frac{q1-q2}{2C}##

That's fine. Once you've factored and grouped the variables for the potential difference expression you should be able to pick out the expression for capacitance in the mix and make the substitution.
 
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  • #28
Ok.
##V##=##\frac{q1-q2}{2sε0}##d

But we know ##C##=##\frac{sε0}{d}##

Here s=A=Area

##\frac{d}{sε0}##=##\frac{1}{C}##

##V##=##\frac{q1-q2}{2C}##

Thanks @gneill.:smile:
 
  • #29
You're welcome!
 

Related to Capacitor Formation with Unequal Charges: A Potential Difference Dilemma

What is a capacitor and how does it work?

A capacitor is an electronic component that stores electrical energy in an electric field. It is composed of two conductive plates separated by an insulating material called a dielectric. When a voltage is applied to a capacitor, it charges the plates and creates an electric field between them. This stored energy can then be released when the capacitor is connected to a circuit.

What are the different types of capacitors?

There are several types of capacitors, including ceramic, electrolytic, film, and variable capacitors. Ceramic capacitors are small, inexpensive, and have a wide range of capacitance values. Electrolytic capacitors have a higher capacitance and are used for filtering and energy storage. Film capacitors are commonly used in high-frequency applications. Variable capacitors have a variable capacitance and are used for tuning circuits.

What factors affect the capacitance of a capacitor?

The capacitance of a capacitor is affected by several factors, including the surface area of the plates, the distance between the plates, and the type of dielectric material used. A larger surface area and a smaller distance between the plates will result in a higher capacitance. The type of dielectric also plays a role, with different materials having different permittivity values that affect the capacitance.

What are some common applications of capacitors?

Capacitors have a wide range of applications in electronics, including power supply filtering, motor starting, timing circuits, and audio amplifiers. They are also used in electronic devices such as cameras, computers, and smartphones. Capacitors are also essential components in electronic circuits for decoupling, coupling, and signal filtering.

What are some common problems with capacitors?

Some common problems with capacitors include leakage, breakdown, and aging. Leakage occurs when the dielectric material breaks down, causing a gradual loss of charge. Breakdown can occur due to excessive voltage or temperature, causing permanent damage to the capacitor. Aging is a natural process that reduces the capacitance of a capacitor over time due to changes in the dielectric material.

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