How do I incorporate electric fields into capacitors?

[MinaciousOviraptor]

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?

 

[Doc Al]

How does electric field relate to voltage? (That's all you need.)

 

[MinaciousOviraptor]

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?

 

[Doc Al]

Read this: Voltage Difference and Electric Field

 

[Doc Al]

Is E-field the same as U?

No. U is energy stored. Read the link I gave above.

 

[YanZhen]

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?

 

[MinaciousOviraptor]

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?

It’s number 5- but that’s literally all it says. I don’t see a relationship to prior problems

 

[YanZhen]

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?

 

[Doc Al]

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.)

 

[YanZhen]

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?

 

[kuruman]

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 $r$.

 

[YanZhen]

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.

 

[Doc Al]

so,how should we describe the relationship between V and E?

Read the link I gave in an earlier post.

 

[YanZhen]

Read the link I gave in an earlier post.

emmm.i can't open it.
could you send me the content?thank you

 

[berkeman]

Read this: Voltage Difference and Electric Field

emmm.i can't open it.
could you send me the content?thank you

You really can't open the Hyperphysics link? I wonder if it's some country blocking thing for your ISP. Here is a snapshot:

 

[kuruman]

I wonder if it's some country blocking thing for your ISP.

Probably the Great Firewall. OP has self-identified as being in China.

 

[YanZhen]

Probably the Great Firewall. OP has self-identified as being in China.

yes

 

[MinaciousOviraptor]

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

 

[Doc Al]

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.

 

[MinaciousOviraptor]

Yes, it's that simple. Doubling the voltage doubles the electric field -- they are proportional.

Thank you!