# Electrostatic potential

1. Apr 24, 2010

### johncena

I can't understand what actually is the electrostatic potential.The definition given in my textbook is as follows : "Electrostatic potential at any point in a region of electrostatic field is the minimum work done in carrying a unit positive charge from infinity to that point without acceleration." But what we mean by saying "two conducters charged to different potentials " and "potential defference across a conducter"?
I also want to know why did we define a quantity called potential ?

2. Apr 24, 2010

### Born2bwire

Potential is related to energy. Specifically, if we have a given potential, then the net work to move from point A to point B is equal to the difference between the value of the potential at those two points, that is, the potential difference. An electrostatic potential is conveniently conservative. So the definition that you have, that potential is the amount of work to bring a charge from infinity to the desired location, works with the potential difference I described above.

Since it is conservative, the net work done to move a charge from point A back to point A is going to be zero, regardless of the path. So the potential to move a charge from infinity to point A is V_a. The potential for a charge from infinity to point B is V_b. If we move from infinity to A to B to infinity again, the net work must be zero since it is conservative. Thus, V_a+V_ab-V_b = 0, thus the work to move a charge from A to B, V_ab, is equal to the difference of the potential between the two points, V_ab = V_b-V_a.

A potential difference across a conductor, which should be zero for statics if the two points are on the same conductor, is the amount of work to move a charge across that conductor. Potential is also useful because the gradient of potential is the force. So if you know the potential of a given system, then you can find the force by taking the gradient.

3. Apr 25, 2010

### johncena

Ohm's law states that the current through a conducter is directly propotional to the potential difference across the two ends of the conducter.
i.e, I $$\alpha$$ V
But how does that imply V is propotional to I ?
Also , I think that it was better to define resistance as I/V. because then resistance is the current flowing per unit potential difference.

4. Apr 26, 2010

### diazona

If I is proportional to V, that means that the equation relating I and V is of the form I = kV, where k is some constant (i.e. something that doesn't depend on V or I). You could rearrange that equation to V = I/k, which is of the same form: now it's V = cI, where c is some constant that doesn't depend on I or V. (You'll note that c is just the reciprocal of k.) That demonstrates that V is proportional to I if I is proportional to V.

Also think about this: if resistance were defined your way, an object with a higher resistance would have more current flowing through it at a given voltage. That doesn't make sense, because we want resistance to measure how much an object resists the flow of electricity through it. There is in fact a quantity called conductance, which is equal to I/V, and which measures how well an object conducts electricity: higher conductance means there will be more current at a given voltage.

5. May 7, 2010

### johncena

Thank you sir....i think i got it now.
But one more doubt , Can we generalise that if 'a' is propotional to 'b','b' is propotional to 'a' ?

6. May 7, 2010

### Staff: Mentor

Yes.
$$a=k\;b$$
implies
$$b=1/k\;a$$

7. May 12, 2010

### johncena

I am very much confused with this statement "When a conducter is given some charge, it's electric potential increases"
I know that potential at a point in an electric field is the work done to move a charge of 1C from infinity to that point without acceleration.But what does it mean by saying 'electric potential of a conducter'?

8. May 12, 2010

### Born2bwire

Should be the amount of work it takes to move a charge from infinity to the conductor. A good conductor is equipotential in addition to the fact that the electric potential is conservative. So it doesn't matter what shape the conductor is or how complicated of a path the charge takes to go from infinity to the surface. All that matters is you bring the charge to the surface of the conductor.

EDIT: I should be specific in saying that this is assuming that our reference zero potential location is at infinity. Often when we talk about problems involving conductors at some non-zero potential, we may have a non-infinity reference for the zero potential. For example, if I have an infinite grounded wire that lies on the axis of a cylindrical shell, we could say that the conducting shell has a potential of 1 V and the wire, being grounded, is 0 V. In this case, which is basically a coaxial cable, we have changed out zero reference to the wire. In fact, because we have an infinite conducting shell at a voltage of 1, the voltage at infinity is infinite. So infinity is not always the reference starting point for the test charge in the definition of potential.

Last edited: May 13, 2010
9. May 13, 2010

### Studiot

We already have a quantity, called conductance, which is defined this way.
It is, of course, the reciprocal of resistance.

Conductance is used in real life calculations and measurements where resistance values are very low.

10. May 13, 2010

### johncena

Thanks for the replies...i think i have got it...

11. May 13, 2010

### johncena

In chemistry, we are studying that in polar molecules like HCl, H possesses a slight +ve charge and Cl possess a slight -ve charge.But how is it possible? As the charge is quantised, how can we say that a particle possess a charge which is less than electronic charge ?

12. May 13, 2010

### Studiot

One explanation is that the shared electron spends slightly more of its time in the region of the chlorine nucleus than in the region of the hydrogen nucleus

13. May 17, 2010

### johncena

(1)While defining a quantity called surface charge density , why we are considering a very small area element (delta S) ? Why we can't define the surface charge density as (total charge)/(total surface area) ?
(2)"As charge can exist only as integral multiple of basic charge (e), therefore, charge distribution is always discrete on account of atomicity of charge"
What is "atomicity of charge"?

Last edited: May 17, 2010
14. May 18, 2010

### Vectronix

I'm surprised nobody mentioned the "electrostatic approximation" already. From what I understand, according to the electrostatic approximation, electric fields are curl-free. This means that the following derivative of the electrostatic field is equal to the zero vector:

$$\nabla\times\textbf{E}=\textbf{0}$$​

In this equation, $$\nabla\times$$ denotes the curl and $$\textbf{E}$$ is the electrostatic field. Since $$\nabla\times\textbf{E}=\textbf{0}$$, that means there exists a scalar field such that

$$\textbf{E}\:=\:-\nabla\varphi$$​

where $$\nabla\varphi$$ denotes the gradient of $$\varphi$$. Therefore the vector field E has a vanishing curl because the curl of a gradient is always zero. The scalar field $$\varphi$$ is called the electrostatic potential because it describes the potential energy that an electric charge will have when placed at a given point in the electrostatic field. So, this is why we define this quantity, because of the electrostatic approximation.

Also, the reason surface area of any surface is partitioned into sufficiently small area elements because the geometry of a general surface changes with position along the surface, so we partition the surface area and use a limiting process so that the largest area element will approach zero, and the smaller the area elements become, the more numerous they are, and the more accurate the value of the total surface area is. In other words, a double integral is used.

I believe the 'atomicity of charge' means that charge is discrete (quantized) because electric charges exist as individual subatomic particles, each one being a multiple of the elementary charge.

Last edited: May 18, 2010
15. May 18, 2010

### johncena

I didn't understand the first paragraph.What is a curl?"electric field are curl free" what is the physical meaning of this statement?

Still I have a large number of doubts in this topic(electrostatics & current)..may be because i am not perfect with the concepts..here are my doubts...
1)Question: Can two equipotential surface intersect each other?Give reasons.
Answer: No. An equipotential surface is always normal to the electric field intensity. If two equipotential surface intersect, then at the point of intersection, there will be two directions for electric field which is not possible.
Doubt: Why an equipotential surface is always normal to the electric field intensity?
2)What is meant by shock? How do we get shock? (biological meaning)
3)Why $$\phi$$ = E.A ?
where $$\phi$$ is the electric flux through a given area, E is the electric field intensity(vector) on that area, and A is the area vector.
4)How is electro negativity measured?
5)What is 1C(coulomb) ? Is it the charge of 6.25*10^(18) electrons? if yes, what is the specialty of the number 6.25*10^(18) ?
6) My physics sir told that when we charge a conducting sphere, charges are distributed uniformly on it's surface and when we charge an insulating sphere, charges are distributed n it's entire volume. Why is it so ?
7)What we mean by a point charge ?

16. May 19, 2010

### Vectronix

Hey, my apologies... I wasn't sure how much of the math you knew so I just decided to post information that I thought might help. Let me explain, then hopefully others will be encouraged to add to what I have said or make corrections as necessary. The curl is a derivative that yields the infinitesimal rotation of a vector field (such as the electrostatic field). It describes how the field lines rotate (if at all). The direction of the curl of a vector-valued function is along the axis of rotation, and the magnitude of the curl of a vector-valued function is the magnitude of rotation. The del operator (sometimes called 'nabla') in Cartesian coordinates is as follows:

$$\nabla\:=\:\frac{\partial}{\partial x}\textbf{i}\:+\:\frac{\partial}{\partial y}\textbf{j}\:+\:\frac{\partial}{\partial z}\textbf{k}$$

$$\partial/\partial x$$ denotes the partial derivative with respect to x, and similarly for y and z. i, j, and k are unit vectors that point toward the positive x-, y-, and z-axes, respectively.

Now, in order to calculate the curl of a vector field such as the electrostatic field, you would evaluate $$\nabla\:\times\:\textbf{E}$$ like the cross product of two vectors (i.e., like the determinant of del and E). The del symbol has no numerical value on its own; it is just a differential operator. If a vector field has a 'vanishing curl' or is 'curl-free' or 'irrotational,' then that just means that its curl is zero and that the field lines do not rotate.

Anyway, since we're talking about static charges here, the field lines from any static source of charge do not revolve around an axis of rotation.

What I was trying to explain before was that since we know that the curl of a static electric field is zero, that means that the electric field (or equivalently, the 'electric field intensity') is defined as the negative gradient of a scalar function. This scalar function describes the 'potential' at any point in the electrostatic field, which describes what the potential energy of a test charge will be when placed in this force field; the negative gradient thereof is a vector quantity, the electrostatic field.

If you could just imagine a set of concentric spheres surrounding a static electric point charge in space, these would be an example of your surfaces of potential. The gradient of any surface function will always point normal to these surfaces when graphed, because of how the gradient is defined. Now, you may ask, "what is between each concentric surface?" The answer is more surfaces that haven't been graphed. If you look at any picture of the electrostatic field surrounding an electric charge, it'll give visual proof that the field lines don't rotate. From this fact, the math says that there should be a function such that the gradient of it will give you the electrostatic field. This is why the potential function is defined in the first place. Then the answer to that question you posted states that it's not possible for the electric field to have two values at one point. This is analogous to the 'vertical line test' to see if a quantity is a function in some of your early math classes. This test fails when one input produces two different outputs; in that case, it isn't a function. I hope this helps. :)

Last edited: May 19, 2010
17. May 19, 2010

### rcgldr

That would be electrical potential energy, not electrical potential. Voltage is a potential: the potential energy per unit charge. The responses in this thread include statements about potential energy as well as potential, which is a bit confusing.

Also that definition where potential or potential energy is zero at infinite distance is appropriate for a field generated by a point source, since potential is relative to 1/r. For an infinite line source, potential is relative to ln(r), and for an infinite plane source, potential is relative to r. For a point source, it makes sense to use ∞ as a reference point, since 1/∞ = 0. For an infinite line source, it makes sense to use 1 as a reference point, since ln(1) = 0. For an infinite plane source, it makes sense to use 0 (the surface of the plane) as a reference point. The field between a pair of unequally charged plates can be treated in the same manner as an infinite plane, if the position being considered is sufficiently away from the edges of the field.

Using gravity as an analogy, a simplifed case that considers the earth to be an infinite plane for relatively small heights above the surface of the earth, gravitational potential energy = m g h, and gravitational potential = g h, both use the surface of the earth as a reference position. For a point source, gravitational potential energy = - G m1 m2 / r, and gravitational potential = - G m / r, which use ∞ as a reference position.