Why is Charge Directly Proportional to Electric Potential?

[Brajesh kedia]

Why is q(charge) directly proportional to v(potential)

 

[Dale]

If you double the charge then you double the potential. Is there some reason you would you expect any other relationship?

 

[Brajesh kedia]

Potential is something what created on a point due to a source charge...what is it exactly first...?

 

[Dale]

What do you already know about the definition of potential? What do you think the answer to your own question is, and what makes you uncomfortable with that answer?

 

[Brajesh kedia]

Potential(kq/r) and is something created on a point due a source charge and if we place a charge at tht point then multiplication of potential wid that chrge make it potential energy

 

[Brajesh kedia]

What is that something created on the point

 

[Dale]

Potential(kq/r) and is something created on a point due a source charge and if we place a charge at tht point then multiplication of potential wid that chrge make it potential energy

Yes, the potential tells you how much potential energy a test charge would have if placed there.

What is that something created on the point

The electromagnetic field.

 

[Brajesh kedia]

And how is q directly proportional to v

 

[Dale]

I already answered this. If you want a different response then you are going to have to provide some more information that will help me understand your question better.

What relationship would you expect? What do you feel was insufficient about the previous answer?

 

[sophiecentaur]

And how is q directly proportional to v

There is plenty of experimental evidence to support this.

 

[Brajesh kedia]

Before that i guess potential differnce confuse me a lot...it is something created in bringing a charge from one point to another point...that means a single charge move say move from point A to B then firstly i would like to ask if it is potential differ at point A or of at B and secondly if potential differnce is created due to change in place of one charge only then why is it said that "potential difference b/w plates"...

 

[Dale]

You can represent the electromagnetic field of a charge in terms of the fields themselves or in terms of the potentials. They are simply two equivalent ways of describing the same thing. As you move charges from A to B you change the fields and therefore you change the potentials also.

I am not sure what you are asking about regarding the plates. Perhaps you are asking about the charges on the plates of a capacitor and the field between the plates.

 

[stedwards]

Why is q(charge) directly proportional to v(potential)

I take it you are asking about the charge and potential at the same coordinates in space and time.

In general, charge is not proportional to electric potential. Given a particular electric potential, several terms contribute to charge. Most vanish if the potential is uniform over space or magnetic fields constant in time.

 

[Brajesh kedia]

Yes am asking so

 

[stedwards]

Measure the electric potential at some given point, \phi_1 due to a bunch of isoloated, static charges, Q1. Replace them with another random bunch Q2 and measure the potential, \phi_2. The resultant potential from the combination Q1+Q2 will be the sum. \phi=\phi_1+\phi_2.

If you double all the charges the potential will double. [Edit: I see Dale has said this, already.]

This is an example of the superpostion principle for electromagnetism. Mathematically, this worked out because electric charge is just a second derivative of potential in this example. The derivative operator is a linear function.

The electric charge--more accurately, the charge density is fully described by the divergence of the electric field (Gauss Law). The electric field, in turn, is the difference of the gradient of the electric potential and the divergence of the magnetic potential. All linear functions.
\rho=\nabla E In terms of potentials, \rho=\nabla \frac{dA}{dt}-\nabla^2 \phi
give or take a few negative signs. \rho is the charge density and A is the magnetic potential.

 

[Brajesh kedia]

Thank u for giving ur time to answer :)