Capacitance and induced charges

In summary: This potential is transferred to the plates, and thus, they are also at the same potential as the 3rd plate?Yes, you are correct. The grounding of the wire connects the two plates together, meaning they have the same potential. And when the third plate is inserted, its charge is distributed between the two plates, making their potentials equal to the potential of the third plate as well.
  • #1
WWCY
479
12

Homework Statement



Two large parallel metal plates are oriented horizontally and separated by a distance 3d. A grounded conducting wire joins them, and initially each plate carries no charge. Now a third identical plate carrying charge Q is inserted between the two plates, parallel to them and located a distance d from the upper plate (a) What induced charge appears on each of the two original plates? (b) What potential difference appears between the middle plate and each of the other plates? Each plate has area A.

Homework Equations

The Attempt at a Solution


I'm not sure how to even attempt the question,

a) I don't know how the distance d is related to charge induced on the top and bottom since E fields between large parallel plates are uniform. What I do think, is that the magnitude of charge induced is equal to Q, by the principle of conservation of charge, but that's all i can come up with.

b) My intuition tells me that using the equation C = ∈A/d and C = Q/V allows me to find the potential difference.

Any help will be greatly appreciated.
 
Physics news on Phys.org
  • #2
You know the induced charge on each plate. You also can calculate the capacitance of each of the capacitors. So V=Q/C gives you the potential difference for each capacitor. What is it that you don't understand?
 
  • #3
ShayanJ said:
You know the induced charge on each plate. You also can calculate the capacitance of each of the capacitors. So V=Q/C gives you the potential difference for each capacitor. What is it that you don't understand?

After obtaining the circuit diagram for this, I'm not quite sure why this is a parallel circuit and why PD across all the 2 capacitors are the same, which is the main assumption i need to solve the problem.

upload_2017-2-1_18-26-21.png


It just doesn't look very parallel-y to me, it actually looks like 2 capacitors in series to me. So what is it that makes this parallel?
 
  • #4
Who said the capacitors are in parallel and who said the potential difference is the same for them?
You're right, they're in series and you still can solve the problem easily. Just look at those formulas you've given in your first post.
 
  • #5
upload_2017-2-1_18-34-10.png


This was pulled from a PDF i downloaded. If this was in series, won't the charge on each of the conducting plates be the same?
 
  • #6
upload_2017-2-1_18-37-17.png


This was the diagram from the pdf, I think it omitted the grounding of the wire. Apologies for not putting this up initially.
 
  • #7
That seems completely wrong to me.
 
  • #8
WWCY said:
It just doesn't look very parallel-y to me
There are two potentials. The capacitors each connect the two potentials, so from that perspective they are parallel.
 
  • #9
haruspex said:
There are two potentials. The capacitors each connect the two potentials, so from that perspective they are parallel.
Could you explain what you meant by 2 potentials?
 
  • #10
When there are only two components in a loop you can view them as being either in series or in parallel (or both!) since the connection topology satisfies the criteria of both.

It's tempting to judge a circuit's topology just by the way the components have been arranged relative to each other on paper, but it's the electrical connections that count.

Here's the same circuit portrayed in two ways simply by rearranging the component positions. All electrical connections (the topology) is exactly the same for each:

upload_2017-2-1_7-31-2.png


There are two nodes (potentials) in the circuit: v0 and v1.
 
  • #11
gneill said:
Here's the same circuit portrayed in two ways simply by rearranging the component positions. All electrical connections (the topology) is exactly the same for each:

View attachment 112365

There are two nodes (potentials) in the circuit: v0 and v1.

Thanks a lot, I think I'm getting the 2 nodes bit. But how exactly does this make V1 = V0 and PD across the 2 capacitors (splitting the middle plate into top and bottom components) identical?

Also, how does the fact that there is a grounding line affect this setup? Or is it simply an means of illustrating that the conductors were initially neutral?

Apologies for being slow with this, I've come to realize Electricity might not be my strongest subject :sorry:
 
Last edited:
  • #12
WWCY said:
Thanks a lot, I think I'm getting the 2 nodes bit. But how exactly does this make V1 = V0 and PD across the 2 capacitors (splitting the middle plate into top and bottom components) identical?
v1 ≠ v0 . They are two separate potentials, and if you look at the diagrams they remain separate and the labels are on the same nodes in both. Here, let me make it clearer by coloring the v0 node:

upload_2017-2-1_7-49-51.png

Starting with the portrayal on the left, try to picture the red wire stretching and the blue wire contracting as the upper capacitor is dragged around and over to left hand side of the diagram, resulting in the portrayal on the right. No connections were changed in the process.
Also, how does the fact that there is a grounding line affect this setup? Or is it simply an means of illustrating that the conductors were initially neutral?
The ground is important because it allows the entry/exit of charge from an external source (ground). It also pins one node at a known potential, since we tend to declare "ground" to be zero potential.
Apologies for being slow with this, I've come to realize Electricity might not be my strongest subject :sorry:
No worries! It's just a new set of "rules and consequences" that you are unfamiliar with. It can be confusing at first as you assimilate them and get used to thinking about circuits and how they behave. It will get easier, and at some point the concepts will "klick" and you'll wonder why it seemed so tricky before!
 
  • #13
gneill said:
The ground is important because it allows the entry/exit of charge from an external source (ground). It also pins one node at a known potential, since we tend to declare "ground" to be zero potential.
So would I be right in saying:
The grounding holds the 2 plates at the same potential, and because these 2 plates are exposed to a conductor of a certain potential (sum of energy associated with all the charge pairs divided by total charge) the PD across the 2 'split' capacitors have to be the same?
 
  • #14
WWCY said:
So would I be right in saying:
The grounding holds the 2 plates at the same potential, and because these 2 plates are exposed to a conductor of a certain potential (sum of energy associated with all the charge pairs divided by total charge) the PD across the 2 'split' capacitors have to be the same?
Sure. You could also say that anything joined by a conducting path (wire) must share the same potential. Hence the labels v0 and v1 on the diagrams.
 
  • #15
gneill said:
Sure. You could also say that anything joined by a conducting path (wire) must share the same potential. Hence the labels v0 and v1 on the diagrams.
Your help cleared things up a lot, many thanks!
 

FAQ: Capacitance and induced charges

1. What is capacitance?

Capacitance is the ability of a system to store an electric charge, measured in farads (F).

2. How is capacitance related to induced charges?

Capacitance is related to induced charges because when a conductor is placed in an electric field, it can become polarized, causing opposite charges to be induced on its surface, creating a capacitor.

3. What is the formula for calculating capacitance?

The formula for calculating capacitance is C = Q/V, where C is capacitance, Q is the charge stored, and V is the voltage.

4. How does the distance between two conductors affect capacitance?

The distance between two conductors directly affects capacitance, as a larger distance between the conductors will result in a lower capacitance and a smaller distance will result in a higher capacitance.

5. What are some real-life applications of capacitance and induced charges?

Capacitance and induced charges have many practical applications, including in electronic circuits, energy storage devices such as batteries and capacitors, and in touchscreens and touch sensors.

Similar threads

Back
Top