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Electric Scalar Potential

Minimum force needed to move charge against E field:

Suppose the E field is constant and points in the –y direction, as shown in the picture here. We can write E = -y^hat*E Suppose further we have a positive test charge q in the field as shown. The field exerts a force on the charge (given as Fe in the figure) equal to Fe = q*E.

The bare minimum force needed to move the positive test charge against the E field is F_ext = -Fe=-qE. Here the subscript “ext” means external...since this is an external force that must be applied. Actually, if we apply this force F_ext then the test charge will not move, since the forces will be perfectly balanced... But if we applied any additional force (even a tiny differential amount), the balance will be broken and the particle will start to accelerate, so we consider F_ext as the minimum force needed to move the positive charge against the E field.

From physics, we define work as the component of force parallel to the direction of motion times the displacement. Hence, the work done (or energy expended) in moving any object a vector differential distance dl while externing a force F_ext is dW=F_ext*dl ...

which in this case, for the positive test charge, equals –qE(dot)dl.

If the test charge is moved a distance dy along y^, then dW = [[[READ EQATION]]]

When we move the test charge a distance dy, we differentially increase it’s potential energy (it’s exactly as if we are raising a ball in the air – we increase it’s gravitational potential energy). We define the differential electric potential energy dW per unit charge the differential electric potential (or differential voltage) dV, that is dV = [[[READ EQUATION]]] dV is numerically equal to the work per unit charge it took for us to move the test charge a differential distance dl against the electric field.

Electric Scalar Potential

V21 is the amount of work per unit charge it takes to carry a positive charge from point P1 to point P2.

Suppose we carry a positive test charge from point P1 to point P2. What is the amount of work per unit charge it takes to carry the test charge against the field from P1 to P2?nt P2. The figure shows various paths we can take to move it from P1 to P2.

Well, using the standard approach from calculus, we view a given path from P1 to P2 as an infinite collection of differential path segments that add up to give the total path.

The potential difference V21 is the total work per unit charge to carry a positive test charge from point P1 to point P2. To find V21,we add up all the differential electric potentials for each of the infinite differential path segments. That is what this integral is doing. In the previous slide we showed that dV = -E(dot)dl, which leads to this line integral.

We commonly use this notation. The potential difference between P2 and P2 is equal to V21 = V2 – V1 = integral from above. If V21 is positive, that means that it takes positive work to move a positive test charge from P1 to P2. This requires a net expenditure of energy on you (the mover of the charge) This energy is put into the electrostatic system in the form of an increase potential energy (it’s analogous to raising a ball up in the air against the force of gravity... Now the ball has more potential energy).

If V21 is negative, that means that it takes negative work to move a positive charge from P1 to P2. This requires you (the mover of the charge) absorb energy. This energy is taken out of the electrostatic system in the form of a decrease of potential energy (it’s analogous to lowering a ball up in the air against the force of gravity... Now the ball has less potential energy).

In electrostatics, it turns out that the potential difference between P2 and P1 is independent of the path. Therefore the electrostatic force is conservative. A conservative force is a force with the property that the work done in moving a particle between two points is independent of the taken path. Gravity is another electrostatic force.

This means that there is zero net work (W) required to move a particle through a trajectory that starts and ends in the same place. Mathematically, this is written as this line integral. The circle over the integral means the path C is closed, that is its initial and final points are the same point. For example a circle is a closed path.

Electric scalar potential due to a point charge.

Given: A point charge Q in free-space.

What is the potential difference V21 = V2 - V1 where P1 is a distance R1 from a point charge and P2 is a distance R2 from a point charge?

[[[prove this equation... For positions on same radial line... And points on different radial lines)

Electric scalar potential due to a point charge (Continued)

In electric circuits, we usually select a convenient node that we call ground and assign it zero reference voltage. In free space and material media, we choose infinity as reference with V = 0.

People simply call V “the electrostatic potential” at position distance R.

From previous slide:

Now set R1 equal to infinity ( ) and omit the subscript 2 in R2

[[[READ RED MESSAGE]]]

So, again is the equation for the potential difference V21 = V2 - V1 where P1 is a distance R1 from a point charge and P2 is a distance R2 . Let’s set R1 to be at infinity. In this case 1/R1 goes to zero. Now, let’s omit the subscript 2 in R2 and we are left with this equation for the potential for a position at distance R from a point charge with respect to infinity. People denote this quantity as V and call it the “electrostatic potential” at position R. Don’t forget what it means. This equation gives the amount of work per unit charge necessary to carry a positive test charge from infinity to a distance R away from the point charge q.

Potential V as a function of distance R from a single point charge q

V is the amount of work (per unit charge) required to carry a positive test charge from infinity to a distance R away from the point charge q.

Electrostatic potential V can be positive or negative. Here is V plotted for a positive point charge. Look how V increases without bound as distance R is decreased.

Imagine yourself pushing a test charge toward the positive point charge. You are the one putting work and effort into this. So you are giving energy to the test charge which you are moving (or rather you are increasing the energy of electrostatic system in general). The plot of V versus R shows that takes more and more work to move a positive test charge closer and closer to the positive point charge.

Ok, now let us switch to the case where point charge is negatively charged. Here is V plotted for a negative point charge. Now there is an attractive force between the negative point charge and the positive test charge. In this case, the negative charge is pulling you along with it. Think of a big dog on a leash that pulls you along with it. The work that you try to do pull back is futile. The system is doing work on you. In this case, your pulling force is in the opposite direction the test charge ends up moving. Thus the work that you do on the charge is negative. In other words, the system has done work on you, thus the electrostatic system has lost potential energy. The plot of V versus R shows that takes more and more negative work to move a positive test charge closer and closer to a negative point charge.

Let’s first bring our attention to the figure on the bottom left, which shows two charged parallel plates. The red lines that point from the positive plate to the negative plate are Electric Field lines. The green dashed lines are called equipotential lines. Equipotential lines are cross sections of equipotential surfaces. An equipotential surface is one in which all points are at the same potential. Note that each green line is labeled with a voltage... 20 V, 15 V, 10 V, 5 V, and 0 V. An equipotential surface must be perpendicular to the E-field at every point. This is because it takes no work to move a charge at right angles to an E-field, so there is no potential differences between two points on a surface perpendicular to the field.

The figure on the bottom right shows two equal but oppositely charged particles next to each other (called an electric dipole). Notice that the electric field lines spring forth from the positive charge (which is a source of E fields) and terminate on the negative charge (which is a sink of E fields). The equipotential lines are drawn again as green dashed lines. Note that they are indeed perpendicular to the E-field lines.

This slide will help you test your understanding of equipotential lines. Which is a large potential V_C or V_D?....

Well, note that they lay on the same equipotential surface, therefore the potential difference between them is zero and V_CD = 0.

Which is a larger potential: V_B or V_C? V_B is larger. - Because point B is closer to the positive charge compared to point C, VB is larger. It would take more work to bring a positive charge from infinity to point B compared to point C.

Which is a larger potential: V_B or V_C? V_B is larger – Again, VB is larger. It would take more work to bring a positive charge from infinity to point B compared to point A.

Relating E to V

Here is an equation we already derived: dV = -E(dot) dl.

Remember, what this is saying. The (negative of the) dot product of E with a differential path length dl is equal to a differential voltage change dV along the direction of dL.

From vector calculus, we recal that the derivative of a scalar field along a direction dl is equal to the gradient of the scalar field (dot) dl. That is shown in this equation here. dV = delta(V) (dot) dl where delta V is the gradient of V.

Given that both equations are true we have delta_V(dot)dl=-E(dot)dl. For this to hold, it must be true that E = -delta(V).

This is a differential relationship between V and E that allows us to determine E for any charge distribution by first calculating V and then taking the negative gradient of V to find E.

Let’s look at an example. Suppose we have a two-dimensional potential plot described by this equation V(x,y) = [[[READ EQUATION]]]. This scalar function is plotted here. The height of this curve at any (x,y) point gives the amplitude of the funciton V(x,y).

We can now take the negative gradient of V to find E. That is done in this equation here. Using Matlab you can easily plot the electric field and the potential on the same figure. Here is a quiver plot of the electric field plotted on top of contour lines describing V.

In case you were interested, here is the MATLAB code I used to generate the previous plot.

By the way, some of the slides in this lecture are tagged with this little picture of the “The more you Know” public service announcement. This tag means the information in this slide is not required material for you to know for the exams and homework, but it’s there simply because it’s good to know.

This is a screenshot of an interactive CD module that you have access to. It allows you to place charges and plot the equipotential lines and electric fields.

Cont.

An electric dipole An electric dipole is two charges, with equal but opposite electric charges, that are separated by a distance.

A dipole can be created, for example, when you place a neutral atom in an electric field, because the positively-charged constituents of the atom will be pulled one way, and the negatively-charged constituents the other way, creating a separation of charge in the direction of the field. Some polar molecules like H20 are naturally polarized and can be modeled as an electric dipole.

Sufficed to say, electric dipoles are everywhere where there is matter, so it’s a good idea to study their fields. Here is a picture of an electric dipole aligned along the z-axis. It consists of a positive charge +q and a negative charge –q separated by a distance d. We are interested in the electrostatic potential V and electric field E at a position R, theta, phi. Notice that we are working in spherical coordinates.

Our strategy will be to first find the electrostatic potential V and then find the electric field (by taking the negative of the gradient of V). We will assume that the distance between charges d is much smaller than the distance of our observation point P.

[[[READ EXAMPLE SOLUTION]]]

(cont.)

[[[READ SOLUTION]]] Note that the electric field drops as 1/R^3 (this is a much faster drop off compared 1/R^2 for a single charge). This is because the further away you are from the electric dipole, the more the positive and negative charges cancel out.

It’s kind of like, suppose you had a yellow LED right next to a blue LED. When you are close to the two LED’s, you can clearly see the two individual colors. When you move far away from the LED’s they start to blend and you would see them as a single greed LED (yellow + blue = green). In the same manner, the further away you get from the electric dipole, the more the positive and negative charges ‘blend’ and you would seem them as a single neutral charge (which produces zero electric field).

P(0,0,h)

Determine the electrostatic potential V at a point P = (0, 0 ,h) along the axis of the ring at a distance h from its center.

In this example, we are asked to find the electrostatic potential V due to a continuous charge distribution. [[[READ and DO Example]]]

V21 = q

4πε0 1 R2 −

1 R1

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V12 = q

4πε0 1 R2 −

1 R1

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R1 →∞

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V = q 4πε0

1 R

4pe

V = q

4πε0 1 R

when q > 0

4pe

when q>0

V = q

4πε0 1 R

when q < 0

4pe

when q<0

ʹ′R

R− ʹ′R

R-