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
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:
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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