11.5: Electrostatic Potential Energy and Potential (2024)

ElectrostaticPotential Energy

Just like forces exist between two objects, the potential energy is always an energy between two objects. In Physics 7Awe tied together the idea of potential energy and force. Welearnedthat the force always points in the direction of decreasing potential energy, and the magnitude of the force depends on the rate of change (slope) of potential energy with distance.This is analogous to releasing a ball on a smooth hillside; the ball starts to roll in the direction of the steepest slope of the hill. In two or three dimensions, the force is the derivative of the potential in the direction of “steepest slope". Going back to the topographical map, contour lines with small distances between them indicate steeper hills. Objects placed here would experience higher acceleration(strongest force) down the hill than objects placed elsewhere.

It is worth repeating here the mathematical description that connects force and potential energy. Most generally, when theforceis the negative of the gradient of the potential:

\[ \overrightarrow F = -\overrightarrow \nabla PE \label{Fgrad}\]

The gradient operator \(\nabla\) is short-hand for writing:

\[ F_{x} = -\frac{dPE}{dx};~~F_{y} = -\frac{dPE}{dy};~~F_{z} = -\frac{dPE}{dz}\]

where x, y, and z are variables for the three spatial dimensions. We will mostly focus on potential energy between objects where the interaction is spherically symmetric(depends on separation \(r\)), such as between point charges. For these cases, Equation \ref{Fgrad} can be written as:

\[ F(r)= -\frac{dPE(r)}{dr}\label{Fr}\]

where \(F(r)\) is the magnitude of a force which points along the radial component \(\hat r\). To solve for potential energy in terms of force, you can rewrite Equation \ref{Fr} in terms of an integralof force over distance. For a force between two point charges described by Coulomb's Law, the electrostatic potential energy is:

\[PE(r)=-\int F(r)dr=-\int\dfrac{kq_1q_2}{r^2}dr=\dfrac{kq_1q_2}{r}+C\label{PEgen}\]

where the constant \(C\) is some constant, and PE has units of Joules. You can check the equation above by taking the derivative of the potential energy with respect to \(r\) and recovering the equation for the force between two charges.Theconstant, \(C\), that appears in the potential energy equation should remind usthat the absolute value of potential energy is arbitrary. The physically relevant quantity isthe change in potential energywhen an object moves from one location to another. For gravitational potential energy, the change in height of the object determinedthe change in potential energy. The actual zero value of the height, and thus potential energy,canchosen at someconvenient location. Forthe Lennard-Jones potential we chose the potential energy to be zero when the two particles were distant and non-interacting. However, this choice of zero did not alter theis the amount of energy required to separate the two particles (break the bond), which is the difference in potential energy between the two particles between far apart and at their equilibrium separation.

Likewise, we canchoose a convenient zero value for the electrostaticpotential energy. When the two particles are far apart, then electric force becomes veryweak. Thus, itmakes sense to choose the potential energy to be zero when the two charges are very far apart, \(r\rightarrow\infty\), as was done for the L-J potential.Taking the limit as the separation goes to infinity in Equation \ref{PEgen}, and setting\(PE(r\rightarrow\infty)=0\), we find that \(C=0\). Thus, we write theelectrostaticpotential energyas:

\[PE(r)=\dfrac{kq_1q_2}{r}\label{PE}\]

Let us think about the connection ofthe potential energy andforce to conceptually understand the equation above. If the two interacting charges are both positive or both negative, then the potential energy is positive. Moving the two charges closer togetherwill result in a positive change in potential energy: \(PE(r)\) increases as \(r\) decreases. The force points in the direction of decreasingpotential energy, thus,moving two like charges closer together is moving them in the opposite direction of the force. On the other hand, ifwe start with two stationary like charges and let them move freely, theywill move apart toward largervalues of \(r\)in the direction of decreasing potential energy. This is consistent with the force for like charges being repulsive and wanting topushthe charges apart. You can also think about this in terms of conservation of energy, if the potential energy decreases, then kinetic energy must increase. Thus, as the two charges repel, then speed up as they separate.

If the two charges have opposite signs, then the potential energy is negative. This means that the potential energydecreases when the two charges move closer together (the potential energy will get more negative). Thus, the two charges will speed up as them come together and their potential energy decreases toward the force which is attractive for two unlike charges.

Another way to think about potential energy is the energy required to bring charges together from infinity (where potential energy is zero). If there are multiple charges, then you need to consider interactions between all the pairs involved. For example, to assemble three charges, \(q_1\),\(q_2\), and \(q_3\), the potential energy is:

\[PE=\dfrac{kq_1q_2}{r_{12}}+\dfrac{kq_1q_3}{r_{13}}+\dfrac{kq_2q_3}{r_{23}}\]

where \(r_{12}\) is the separation between charges 1 and 2, \(r_{13}\) is the separation between charges 1 and 3, and so on.

Electrostatic Potential

In the previous section we defined the electric field, which is a vector field generated by a charge, acollection of source charges, or a macroscopic charged object. Once the electric field is determined, one can compute theforce that a test charge feels when it is placed in theelectric field. We can define an analogous field related to potential energy. Let usassign a charge or a collection of charges that generates a scalar field defined everywhere in space, which we call the electrostaticpotential, V, sometimes referred to as simply electric potential or voltage.Then, we can analyze that amount of potential energy required to to bring a test electric charge to some location in thispotential field. For one point sourcecharge, \(q_{\text{source}}\), the electric potential is defined as:

\[V_{q_{\text{source}}}=\dfrac{kq_{\text{source}}}{r}\label{V}\]

The units of the electric potential are joules per coulomb, \(\dfrac{J}{C}\), also known as Volts, V. Theamount of potential energy required to bringa test charge,\(q_{\text{test}}\), from infinity to adistance of \(r\)from the source chargeis then given by:

\[PE=q_{\text{test}}V_{q_{\text{source}}}\]

Although, there is a direct relationship between force, which is a vector quantity, and energy, the potential energy is a scalar quantity which has no direction. Since the potential energy is a scalar, so is the electric potential based on equation above.

The same rules of superposition apply to the electric potential as did for the electric field. If there is a collection of chargesthen the total electric potential generated by these charges is the sum of the electric potentials due to each charge. Exceptthe total electric potential is much easier to calculate since it is a scalar field, thus it is simplya sum of numbers. The only thing that you need to be careful about is the sign of each charge since the potential can be either positive or negative. For \(N\) charges, the total electric potential:

\[V_{\text{tot}}=\sum_{i=1}^{N}\dfrac{kq_i}{r_i}\]

where \(r_i\) is the distance from the location where the voltage is calculated to the charge \(q_i\).

Summary:Force,Field,Potential, and Potential Energy

In the previous section we have established the connection between electric force and field: source charges generatea vectorfield, and a test charge feels a force in the presence of the field. Likewise, source charges generates apotential scalar field, and a test charge placedin this field will havesome potential energy. Mathematically, theserelationships are given in the following way:

\[\vec F=q_{\text{test}}\vec E_{q_{\text{source}}}\]

\[PE=q_{\text{test}}V_{q_{\text{source}}}\]

The quantities on the left side of the two equations above are depend on both the source and the test charge since they describe the interaction. The quantities on the right side of the equation that multiply the test charge solely depend on the source charges.

We also used the previously learnedrelationship between force and potential energy, which leads directly to a new relationship between electric field and electric potential.For a point charge these are:

\[ \vecF = -\dfrac{dPE}{dr}\hat r\]

\[ \vec E= -\dfrac{dV}{dr}\hat r\label{EV}\]

The second equation comes directly from the first by dividing each side of the first equation by test charge: dividingthe left side of the equation by \(q_{\text{test}}\) weget the electric field, and dividingthe right side of the equation by \(q_{\text{test}}\) we getthe electric potential. From the equation above we conclude that electric field points in the direction of decreasing potential, since force points in the direction of decreasing potential energy.

Here is the summary of the important concepts to keep in mind while working with these four quantities:

• An electric field is generated by a source charge or a collection of source charges.
• An electric field points away from a positive charge and toward a negative charge.
• For a collection of source charges, the total electric field atsomelocation is computed by adding the electric field vectors generated by eachcharge.
• A positive test charge feels a force in the same direction as the electric field, and anegative test charge feels a force in the opposite direction of the electric field.
• An electric potentialis generated by a source charge or a collection of source charges.
• An electric potential is always positive and decreases away from a positive charge, and is always negative and increases away from a negative charge.
• For a collection of source charges, the total electric potential at somelocation is computed by adding the electricpotentialsgenerated by eachcharge.
• A positive test charge decreasesin potential energy as it moves toward decreasingelectric potential, and a negativetest charge increasesin potential energy as it moves toward decreasingelectric potential.
• Electric field points in the direction of decreasing electric potential.
• Electric force points in the direction of decreasing potential energy.

The figure below demonstrates some of these common misconceptions inconnecting the electric field and potential. The scenario on the left shows a positive test charge which is moved along the line in between two identical positive source charges. The situation on the right shows a positive test charge that is moved between a positive and a negative source charge of equal magnitude.

Figure 11.5.1: Field and Potential For Two Point Charges

In the scenario on the left, the force is zero on the test charge when it is right in between the two charges.Thetotal electric fieldis zero at that location,since the electric fields from each sourcecharge are equal in magnitude but opposite in direction.However, this does not imply that the potential is zero at that location. In fact, the total potential has an equal positive contribution from each charge, resulting in a total potential of \(V_t=2\dfrac{kq}{r_1}\). You may thinkhow can the electric field be zero and the potential be non-zero based on their relationship given by Equation \ref{EV}. The reason that the field goes to zero, is that at the central location the potential is not changing for very small displacements along the central line. Right above and right below the central location the distance to both charges is the same. Thus, thepotential is constant for a small vertical displacement through the center resulting is zero rate of change of the electric potential, and thus zero electric field at the center.As the test chargemoves up and away from the two charges, you can see that there is a net vertical component to the total electic field, so the charge will feel a vertical force. The electric potential decreases in that direction since \(r_2>r_1\), which is consistent with the electric field pointing in the direction of the decreasing potential.

The scenario on the right is another interesting situation, sincethe electric potential is zero along the line separating the two charges. This is the case since the distance to each charge is the same, but one is positive and the other is negative of equal magnitude, so the total potential adds to zero. However, this does not necessarily mean thatthe electric field is zero along that line. In fact as shown in the figure, there electric field is non-zero everywhere along that line and points to left, toward the negative charge and away from the positive. At first glance thisdoes not seem to be consistent with Equation \ref{EV} since the potential is not changing, but yet the field is non-zero. However, we need to remember that the field is a vector quantity.Equation \ref{EV} tells us that the field must be zero along the direction where the potential is not changing. This is true for this case, since the field has novertical component. However, thefield can still be non-zero in a direction perpendicular to the central line, as seen in the figure. In fact, if you calculated the potential along the direction of the field, you would discover that it does indeed decrease in that direction since it's getting closer to the negative charge and further from the positive one.

Equipotential Surfaces

In the previous section we describe a useful tool for depicting electric fields using field lines. A similar informativedepiction exists for the electric potential. Going back to topographical maps in Section 11.1, we saw that itdescribed a scalar field with lines representing location of constant value of the field (elevation). The elevation difference between neighboring lines was fixed. This means thatregions where the contour lines were most dense, where regions where the slope was the largest (elevation changed mostrapidly with distance). The oppositewas true for regions of low density.

If we want to use a similar representation for electric potential, we will draw curves where the potential is constant, known as equipotential surfaces.Equation \ref{V} tells us that theelectric potential is constant along a radius, \(r\), around a point charge. Therefore, equipotential surfacesaroundpoint charges will be spherical shells. In a two-dimensional illustration, the surfaces will become circles around the charges.

As fora topographical map where neighboringcontour lines were separated by a fixedvalue of elevation, we want to drawequipotential surfaces separated by a fixed value of potential. Thus, we need to determine how voltage changes with distance in order to know how to space the equipotentials. The figure below is a plot of electric potentialas a function of distance for a positive source charge.

Figure 11.5.2: Electric Potential Plot for a Positive Charge

From the plot you can see that for small values of \(r\) (close to the source charge), the voltage changes more rapidly (slope is larger)compared to larger distances from the charge. Therefore, two equipotential surfaces nearthe charge will be separated by asmaller distance, \(\Delta r\), compared to neighboring equipotential surfaces further from the charge, \(\Delta r\) is bigger for the same \(\Delta V\).

For a positive source charge the electric potential is positive and decreases with distance. For a negative source chargethe electric potential is negative and increase with distance. All of these concepts are illustrated in the figure below, together with the field lines for a positive and a negative point charge. As shown the equipotential surfaces get less dense away from the change. The electric potential values are arbitrarily selected to change by1V, for the sake of demonstrating the sign and change of the potential.

Figure 11.5.3: Equipotential Lines for Point Charges

The field lines appear to beperpendicular to the equipotential surfaces and point in the direction of decreasing potential. Since the electric field is the rate of change of the electric potential, when the electric potential is constant the electric field must be zero according to Equation \ref{EV}. Therefore, since equipotentialsdescribe surfaces of constant potentialthere cannot be any component of the electric field parallel to the equipotential. Therefore, fields lines are always perpendicular to equipotential surfaces.In addition, the density of equipotential lines correlateswith the density of field lines. The electric field has the greatest magnitude where the the field lines are most dense. Larger magnitude of electric field, implies larger derivativeof voltage with distance according toEquation \ref{EV}. Thus,the equipotential surfaces will be densestwhere the field lines are most dense.

In the previous section we looked as a special case of a charged infinite conducting plane, where we discovered that the electric field is constant with distance away from the plane. For this scenarioEquation \ref{EV} tells us that the electric potential has to be linear with distance in order to result in a constant electric field, sincethe derivativeof a linear function results in a constant. In other words, the change in electric potentiallinearly depends on the distanceaway from the plane:

\[\Delta V_{\text{plane}}=-E\Delta r\label{Vplane}\]

Therefore, the equipotential surfaces away from the plane will be equidistant,as shown in the figure below. This is consistent with the fact that the density of the electric field lines is constant, implying that the density of the equipotential surfaces must be constant as well. As previously argued, the equipotential lines areperpendicular to the field lines.

Figure 11.5.4: Equipotential Lines for an Infinite Charged Plane

Arbitrary values of the electric potential are assigned in the figure above to stress the potentialis positive anddecreasingaway from a positively charged plane in the direction of the electric field. A positive test charge placed in this field will then will a constant forceaway from the plane,and its potential energy will decrease as it moves away from the plane. The opposite is true fo a negative test charge, it will feel a force toward the plane, and its potential energy will decreasetoward the plane,in the direction of increasing electric potential.