How do I incorporate electric fields into capacitors?

In summary, the conversation discusses the relationship between voltage and electric field, specifically in the context of a capacitor. The main takeaway is that doubling the voltage will also double the electric field, as they are directly proportional. The conversation also touches on the difference between U (energy stored) and electric field, and the incorrect equation U=(1/2)QV is mentioned.
  • #1
MinaciousOviraptor
5
2
Homework Statement
A capacitor with a value of C stores a charge Q and has an electric field E between the plates when a voltage V is across it. What will the new value of the electric field be if:

a) the voltage is 2V
b) the voltage is 4V
Relevant Equations
U=(1/2)CV^2
U=(1/2)QV
U= (1/2)Q^2V
Q=CV
I know that I’m supposed to use proportional reasoning, but where does electric field even fit in? For whatever equation, I know I’m supposed to see how increasing the voltage by either 2 and 4 volts related to electric field. If electric field is the same as “U”, then wouldn’t it be U=(1/2)*(1)*(3V) using the equation U=(1/2)QV?
 
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  • #2
How does electric field relate to voltage? (That's all you need.)
 
  • #3
Doc Al said:
How does electric field relate to voltage? (That's all you need.)
But what’s the relevant equation? Is E-field the same as U?
 
  • #5
MinaciousOviraptor said:
Is E-field the same as U?
No. U is energy stored. Read the link I gave above.
 
  • #6
MinaciousOviraptor said:
Homework Statement:: A capacitor with a value of C stores a charge Q and has an electric field E between the plates when a voltage V is across it. What will the new value of the electric field be if:

a) the voltage is 2V
b) the voltage is 4V
Relevant Equations:: U=(1/2)CV^2
U=(1/2)QV
U= (1/2)Q^2V
Q=CV

I know that I’m supposed to use proportional reasoning, but where does electric field even fit in? For whatever equation, I know I’m supposed to see how increasing the voltage by either 2 and 4 volts related to electric field. If electric field is the same as “U”, then wouldn’t it be U=(1/2)*(1)*(3V) using the equation U=(1/2)QV?
Do you have original question?
Why U=(1/2)QV and U=(1/2)Q^2V?
(1/2)QV≠(1/2)Q^2V.I can't understand it.
Could you explain it?
 
  • #7
YanZhen said:
Do you have original question?
Why U=(1/2)QV and U=(1/2)Q^2V?
(1/2)QV≠(1/2)Q^2V.I can't understand it.
Could you explain it?
2B5A080D-3B47-49A5-ABC8-F3B0AE1A106B.jpeg
It’s number 5- but that’s literally all it says. I don’t see a relationship to prior problems
 
  • #8
E=U/d
d is a constant
so Ea=2E Eb=4E
and U is like mgh,E is like g.one in the electric field,one in the force field.
understand?
 
  • #9
MinaciousOviraptor said:
Relevant Equations:: U=(1/2)CV^2
U=(1/2)QV
U= (1/2)Q^2V
Q=CV
The highlighted equation is incorrect.

As I stated before, you won't need any of these. Just the relationship between V and E. (Read the link I gave.)
 
  • #10
C=Q/V C is a constant
E=F/q=((kQq)/(d^2))/q=kQ/(d^2)=kCV/(d^2)
is this the answer you want?
 
  • #11
YanZhen said:
C=Q/V C is a constant
E=F/q=((kQq)/(d^2))/q=kQ/(d^2)=kCV/(d^2)
is this the answer you want?
This is misleading and wrong. You are misleading the OP by asserting that the total charges ##\pm Q##, which are distributed uniformly over the plates, are point charges. It is wrong because the field due to two point charges is the superposition of two ##1/r## fields while the field between the capacitor plates is uniform and does not depend on ##r##.
 
  • #12
kuruman said:
This is misleading and wrong. You are misleading the OP by asserting that the total charges ##\pm Q##, which are distributed uniformly over the plates, are point charges. It is wrong because the field due to two point charges is the superposition of two ##1/r## fields while the field between the capacitor plates is uniform and does not depend on ##r##.
so,how should we describe the relationship between V and E?
i'm so out of ideas.
 
  • #13
YanZhen said:
so,how should we describe the relationship between V and E?
Read the link I gave in an earlier post.
 
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  • #14
Doc Al said:
Read the link I gave in an earlier post.
emmm.i can't open it.
could you send me the content?thank you:biggrin:
 
  • #15
Doc Al said:
Read this: Voltage Difference and Electric Field
YanZhen said:
emmm.i can't open it.
could you send me the content?thank you:biggrin:
You really can't open the Hyperphysics link? I wonder if it's some country blocking thing for your ISP. Here is a snapshot:

1666621988470.png
 
  • #16
berkeman said:
I wonder if it's some country blocking thing for your ISP.
Probably the Great Firewall. OP has self-identified as being in China.
 
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  • #17
kuruman said:
Probably the Great Firewall. OP has self-identified as being in China.
yes
 
  • #18
Doc Al said:
Read the link I gave in an earlier post.
So I’m disregarding Capacitance and Charge and just focusing on the relationship between V and E—> V=ED. If V=2V then E would also be 2
 
  • #19
MinaciousOviraptor said:
So I’m disregarding Capacitance and Charge and just focusing on the relationship between V and E—> V=ED. If V=2V then E would also be 2
Yes, it's that simple. Doubling the voltage doubles the electric field -- they are proportional.
 
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  • #20
Doc Al said:
Yes, it's that simple. Doubling the voltage doubles the electric field -- they are proportional.
Thank you!
 
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FAQ: How do I incorporate electric fields into capacitors?

How do electric fields affect the capacitance of a capacitor?

Electric fields play a crucial role in determining the capacitance of a capacitor. The greater the electric field between the plates of a capacitor, the higher the capacitance will be. This is because the electric field creates a force that pulls the opposite charges on the plates closer together, increasing the amount of charge that can be stored on the plates.

Can the electric field strength in a capacitor be changed?

Yes, the electric field strength in a capacitor can be changed by altering the distance between the plates or by changing the amount of charge on the plates. Increasing the distance between the plates decreases the electric field strength, while increasing the charge on the plates increases the electric field strength.

How do I calculate the electric field strength in a capacitor?

The electric field strength in a capacitor can be calculated by dividing the voltage across the plates by the distance between the plates. This can be represented mathematically as E = V/d, where E is the electric field strength, V is the voltage, and d is the distance between the plates.

What is the relationship between electric field and potential difference in a capacitor?

The electric field and potential difference in a capacitor are directly proportional. This means that as the electric field increases, the potential difference between the plates also increases. This relationship can be expressed as E ∝ V, where E is the electric field and V is the potential difference.

How do I incorporate electric fields into the design of a capacitor?

Incorporating electric fields into the design of a capacitor involves considering factors such as the distance between the plates, the material used for the plates, and the dielectric material between the plates. By manipulating these factors, the electric field strength and capacitance of the capacitor can be adjusted to suit specific needs and applications.

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