Chapter

Potential and Potential Energy

Blinn College - Physics 2326 - Terry Honan

. - Electric Potential Energy

Review of Potential Energy

Recall from the first semester course that a force is defined to be conservative when its work, W=∫ (OverVector[F])·d(OverVector[r]), is zero for any closed path.

0=∮ (OverVector[F])·d(OverVector[r])     ((OverVector[F]) is conservative.)

The circle in the above integral symbol denotes the path is closed. An equivalent definition of a conservative force is that work is independent of path, meaning that the work is the same for all paths between the same two endpoints.

In mechanics, energy is introduced through the work-energy theorem: (Subscript[W, net])=Δ K, where K=(1/2)m (v^2) is defined as the kinetic energy. When a force is conservative we can write its work in terms of the endpoints of the path only. We define potential energy for a conservative force by

Δ U=-W=-∫ (OverVector[F])·d(OverVector[r]) .

The reason for the negative sign is to get a term to add to the kinetic energy in the work-energy theorem, moving the term from the work side to the energy side of the equation gives the minus sign.

Examples of conservative forces from the first semester course are gravity, both for a uniform field and for a varying field, and the elastic force of a spring. We will now see that the electrostatic force is conservative; this will allow the definition of electric potential energy.

Potential Energy for Two Point Charges

We want to find an expression for the potential energy between two point charges, (Subscript[Q, 1]) and (Subscript[Q, 2]). First we must find the work done by the electrostatic force and show that that force is conservative. To do this we will begin with (Subscript[Q, 1]) in a fixed position at the origin and move (Subscript[Q, 2]) along a path from (Subscript[(OverVector[r]), i]) to (Subscript[(OverVector[r]), f]).

To find the work to move (Subscript[Q, 2]) we integrate the Coulomb’s law force on (Subscript[Q, 2]) along a path W=∫ (OverVector[F])·d(OverVector[r]). The force on (Subscript[Q, 2]) due to (Subscript[Q, 1]) is (OverVector[F])=(Subscript[k, e])(Subscript[Q, 1]) (Subscript[Q, 2]) ((OverHat[r])/(r^2)), where (OverVector[r]) is the vector from (Subscript[Q, 1]) to (Subscript[Q, 2]). The work integral becomes

W=∫(OverVector[F])·d(OverVector[r])=(Subscript[k, e]) (Subscript[Q, 1]) (Subscript[Q, 2]) (∫_( (Subscript[(OverVector[r]), i]))^( (Subscript[(OverVector[r]), f])))((OverHat[r])/(r^2))·d(OverVector[r])

Multiplying a vector by a unit vector gives the vector component in the direction of the unit vector, so (OverHat[r])·d(OverVector[r]) is just the radial component of d(OverVector[r]) which is just dr=d‖(OverVector[r])‖. This is the change in the radial distance from the (Subscript[Q, 1]) at the origin: dr=‖(OverVector[r])+d(OverVector[r])‖-‖(OverVector[r])‖ and note that it is different from the magnitude of d(OverVector[r]).

(OverHat[r])·d(OverVector[r])=dr  where  dr=d‖ (OverVector[r]) ‖=‖ (OverVector[r])+d(OverVector[r]) ‖-‖ (OverVector[r]) ‖≠‖ d(OverVector[r]) ‖

With the substitution of dr for (OverHat[r])·d(OverVector[r]), the limits of the integral depend only on the radial distances of the endpoints. The integral along a contour has become a single variable integral.

W=(Subscript[k, e]) (Subscript[Q, 1]) (Subscript[Q, 2]) (∫_( (Subscript[r, i]))^( (Subscript[r, f])))((dr)/(r^2))=-(Subscript[k, e]) (Subscript[Q, 1]) (Subscript[Q, 2]) ((1/(Subscript[r, f]))-(1/(Subscript[r, i]))),

Clearly this integral is the same for any two paths between the same two endpoints, so the force is conservative and we can then define the potential energy.

Δ U=-W=(Subscript[k, e]) (Subscript[Q, 1]) (Subscript[Q, 2]) ((1/(Subscript[r, f]))-(1/(Subscript[r, i]))),

To go from Δ U to an expression for U, we need to find a function U(r), to describe the potential energy as a function of r, the distance between the charges.  It must satisfy ΔU=U((Subscript[r, f]))-U((Subscript[r, i])) and is unique up to an arbitrary constant.  The simplest choice is

U=(Subscript[k, e]) (((Subscript[Q, 1]) (Subscript[Q, 2]))/r)

where the arbitrary constant is chosen to make the potential energy zero at infinity.

U(∞)=0   or more precisely   (lim_(r->∞))U(r)=0

Potential Energy of Charge Distributions

For a three charge configuration we have potential energy between each pair.  Taking the distance between charges (Subscript[Q, i]) and (Subscript[Q, j]) to be (Subscript[r, i,j]) gives the potential energy as

U=(Subscript[k, e]) (((Subscript[Q, 1]) (Subscript[Q, 2]))/(Subscript[r, 1,2]))+(Subscript[k, e]) (((Subscript[Q, 1]) (Subscript[Q, 3]))/(Subscript[r, 1,3]))+(Subscript[k, e]) (((Subscript[Q, 2]) (Subscript[Q, 3]))/(Subscript[r, 2,3])).

The general expression is

U=(Subscript[k, e])(∑_(i<j)) (((Subscript[Q, i]) (Subscript[Q, j]))/(Subscript[r, i,j])),

where the sum is over all pairs of charges. Keeping i<j avoids double counting and excludes i=j, which would correspond to the energy between a particle and itself.

Example . - Energy and Electrons Released from Rest

Solution

({{({{(Subscript[r, i])=5.0×(10^(-10))(QuantityUnit)
e=1.602×(10^(-19)) C }, {m=(Subscript[m, electron])=9.11×(10^(-31))kg}})}})

K=2×(1/2)m (v^2)

U=(Subscript[k, e]) (((Subscript[q, 1])(Subscript[q, 2]))/r)=(Subscript[k, e]) (((-e) (-e))/r)=(Subscript[k, e]) ((e^2)/r)

(Subscript[K, i])+(Subscript[U, i])=(Subscript[K, f])+(Subscript[U, f])  ⟹  0+(Subscript[k, e]) ((e^2)/(Subscript[r, i]))=2×(1/2)m (Subscript[v, f]^2)+0  ⟹  (Subscript[v, f])=(Sqrt[(((Subscript[k, e])(e^2))/(m (Subscript[r, i])))])=712000(((QuantityUnit))/((QuantityUnit)))

Example . - Work to Move a Charge - Potential Energy Method

Solution

U=(Subscript[k, e])(∑_(i<j)) (((Subscript[Q, i]) (Subscript[Q, j]))/(Subscript[r, i,j])),

(Subscript[U, i])=4×(Subscript[k, e])((-(q^2))/a)+2×(Subscript[k, e])((q^2)/((Sqrt[2]) a))=-(4-(Sqrt[2]))(Subscript[k, e])((q^2)/a)=-2.5858 (Subscript[k, e])((q^2)/a)

(Subscript[U, f])=2×(Subscript[k, e])((-(q^2))/a)+(Subscript[k, e])((q^2)/((Sqrt[2]) a))=-(2-(1/(Sqrt[2])))((q^2)/a)=- 1.2929((q^2)/a)

W= Δ U = (Subscript[U, f])-(Subscript[U, i])=(2-(1/(Sqrt[2])))((q^2)/a)=1.2929((q^2)/a)

. - Definition of Electric Potential

Basic Definitions

First some terminology:  We will define electric potential V as the potential energy per test charge, V=U/(Subscript[q, 0]). The similarity of the letters U and V combined with the similar names, potential energy versus potential, creates confusion between these related but distinct notions. Voltage is potential difference Δ V; that is, differences in potential. When the sign of Δ V is ignored, we will sometimes use the standard, although somewhat ambiguous, notation where voltage is written as just V, instead of the more precise |Δ V|.

The diagram shows the underlying relationships between the force, field, potential energy and potential. The potential V, at the lower right, is what we are introducing here.

({{(OverVector[F]), ⟵, (OverVector[E])=((OverVector[F])/(Subscript[q, 0])) and  (OverVector[F])=Q (OverVector[E]), ⟶, (OverVector[E])}, {↑,          , ({{ }, { }}) ,  , ↑}, {Δ U=-∫ (OverVector[F])·d(OverVector[r]),          ,  ,          , Δ V=-∫ (OverVector[E])·d(OverVector[r])}, {↓,         ,  ({{ }, { }})  ,  , ↓}, {U, ⟵, V=(U/(Subscript[q, 0])) and  Δ U=Q Δ V, ⟶, V}})

Recall that we define the electric field by considering the force on a test charge and by dividing the charge into the force,  (OverVector[E])=(OverVector[F])/(Subscript[q, 0]).  The potential is defined from the potential energy similarly.  The potential is defined as the potential energy per test charge.

V=(U/(Subscript[q, 0]))

The zero of potential energy is arbitrary and so is the zero of potential. When a charge Q is moved across a potential difference Δ V we get a change in potential energy given by

Δ U=Q Δ V.

If the above expressions are taken as the definition of V then dividing both sides of Δ U=-∫ (OverVector[F])·d(OverVector[r]) gives an expression for the potential difference as an integral of the electric field over a contour

ΔV=-∫(OverVector[E])·d(OverVector[r]).

This is the fundamental expression showing how potential can be found from electric fields.

A Physical Example and Gravitational Analog

Consider the 12(QuantityUnit)  battery in a car connected across some load labeled as “Stuff”: the starter, headlights, car stereo, etc.  Charge flows from the positive terminals to negative terminals, where when we discuss the flow of charge we mean the flow of positive charge.  When a positive charge q flows from the positive terminal to the negative terminal the potential difference (voltage) is negative

Δ V=(Subscript[V, f])-(Subscript[V, i])=(SubMinus[V])-(SubPlus[V])=-12V.

It follows that since the charge q is positive, the change in the potential energy Δ U is negative.

q>0   ⟹  Δ U=q Δ V=q×(-12V)<0

Here U is the electrical energy stored in the battery.  This is decreasing and that is providing energy to the load; it turns the starter, powers the headlights or powers the car stereo.

There is a very good gravitational analogy for this. Consider water in an elevated tank, where the energy of flowing water is utilized before flowing into a lower tank. This could be water flowing over a paddle wheel before spilling into a lower tank or some more efficient means of harnessing this energy.  If h is the (positive) height difference between the water levels of the two tanks, then when a mass m flows from the top to the bottom tank, where Δ y=-h, there is a drop in potential energy.

Δ U=m g Δ y=-m g h<0

A (theoretical) maximum of m g h of energy can then be harnessed for any application.

In both cases we are losing potential energy; the battery and its gravitational analog are being depleted.  To replenish the energy of the gravitational setup we may mechanically pump water from the bottom to the top, where the external work of the pump adds to the potential energy.  Moving a mass m of water from the bottom to the top gives

(Subscript[W, external])=Δ U=m g Δ y=+m g h>0.

The alternator of a car is a type of a generator; it converts external work provided by the car’s engine into electrical energy by pushing (positive) charge to higher potential.

(Subscript[W, external])=Δ U=q Δ V=q×(+12V)>0

Units:  The SI unit for Electric Potential and Voltage is: volt=(QuantityUnit)=(QuantityUnit)/(QuantityUnit)

Example . - Accelerating Charge

Solution

({{({{V=|Δ V|=12(QuantityUnit)
e=1.602×(10^(-19)) C }, {m=(Subscript[m, electron])=9.11×(10^(-31))kg}})}})

(Subscript[K, i])+(Subscript[U, i])=(Subscript[K, f])+(Subscript[U, f])  ⟹  0=Δ K+Δ U  ⟹ |Δ K|=|Δ U|

|Δ K|=|Δ U|  ⟹  (1/2)m (Subscript[v, f]^2)=e V  ⟹  (Subscript[v, f])=(Sqrt[((2 e V)/m)])=2.05×(10^6)(((QuantityUnit))/((QuantityUnit)))

Solution

. - Potential for Charge Distributions

We have found expressions for electric fields due to charge distributions, first considering the point charge and then the cases of discrete and continuous distributions. We will now do the same for potentials. Since potential is a scalar we will see that the potential calculations are simpler because they lack the vector complications in the field calculations.

Point Charge

We defined the electric field as the force per test charge, (OverVector[E])=(OverVector[F])/(Subscript[q, 0]). When we found the field at some point P due to a point charge Q we uses Coulomb’s law to find the force of Q on some test charge (Subscript[q, 0]) at P and then divided by the test charge. We will do the analogous thing here to find the electric potential V at the point P due to a point charge using the definition of the electric potential as the potential energy per test charge: V=U/(Subscript[q, 0]).

The potential energy on a test charge (Subscript[q, 0]) at P is U=(Subscript[k, e])Q (Subscript[q, 0])/r. Dividing by (Subscript[q, 0]) we get the electric potential

V(r)=(Subscript[k, e])(Q/r)

Just as the potential energy goes to zero at infinity, so does the potential.

V(∞)=0  or more precisely  (lim_(r->∞))V(r)=0.

Discrete Distribution

To find the electric potential at a point P due to charges (Subscript[Q, i]) we will take (Subscript[r, i]) to be the distance from (Subscript[Q, i]) to P.  The potential is

V=(Subscript[k, e])(∑_i)((Subscript[Q, i])/(Subscript[r, i]))

Example . - Two Point Charges (Continued)

Solution

V=(Subscript[k, e])((Subscript[Q, 1])/(Subscript[r, 1]))+(Subscript[k, e])((Subscript[Q, 2])/(Subscript[r, 2]))

(Subscript[V, 1])=(Subscript[k, e])((Subscript[Q, 1])/(Subscript[r, 1]))=9.0×(10^9) ((N·(m^2))/(C^2))·((80×(10^(-6))C)/(4 m))=180000(QuantityUnit)

(Subscript[V, 2])=(Subscript[k, e])((Subscript[Q, 2])/(Subscript[r, 2]))=9.0×(10^9) ((N·(m^2))/(C^2))·((-50×(10^(-6))C)/((Sqrt[(1^2)+(2^2)]) m))=-201250(QuantityUnit)

V=(Subscript[V, 1])+(Subscript[V, 2])=-21200(QuantityUnit)

Example . - Work to Move a Charge - Electric Potential Method

Solution

W=Δ U=q Δ V

Δ V=(Subscript[V, f])-(Subscript[V, i])=(Subscript[V, ∞])-(Subscript[V, P])=0-(Subscript[V, P])

V=(Subscript[k, e])(∑_i)((Subscript[Q, i])/(Subscript[r, i]))=(Subscript[k, e])((-q)/a)+(Subscript[k, e])(q/((Sqrt[2])a))+(Subscript[k, e])((-q)/a)=-(Subscript[k, e])(q/a)(2-(1/(Sqrt[2])))=-1.293  (Subscript[k, e]) (q/a)

W=Δ U=q Δ V=-q (Subscript[V, P])=(Subscript[k, e])((q^2)/a)(2-(1/(Sqrt[2])))=1.293  (Subscript[k, e]) ((q^2)/a)

Continuous Distribution

As was done in the case of the electric field we will take r to be the distance from dq to P.  The potential is

V=(Subscript[k, e])∫((dq)/r).

Using Integrals to Calculate Electric Potentials

Mathematically break up the distribution into an infinite number of infinitesimal pieces.  This requires choosing an integration variable and its limits.

Find dq, the infinitesimal charge of each infinitesimal piece.  You must write the dq in terms of the differential d of the integration variable.

Write an expression for r, the distance from dq to the point P where the value of the field is wanted.  Note that r is the magnitude of the vector (OverVector[r]) needed to find the electric field.

Put the pieces together to form a properly-defined definite integral and evaluate the integral.

Example . - Uniform Line of Charge

Solution

V=(Subscript[k, e]) ∫((dq)/r)=(Subscript[k, e]) (∫_(-L/2)^(L/2)) (1/((Subscript[x, 0])-x))(Q/L)dx=(Subscript[k, e]) (Q/L)(∫_(-L/2)^(L/2)) ((dx)/((Subscript[x, 0])-x))

(Subscript[(Subscript[V=-(Subscript[k, e]) (Q/L)ln((Subscript[x, 0])-x)|, -L/2]^(L/2))=(Subscript[k, e]) (Q/L)ln((Subscript[x, 0])-x)|, L/2]^(-L/2))=(Subscript[k, e]) (Q/L)(ln((Subscript[x, 0])+L/2)-ln((Subscript[x, 0])-L/2))=(Subscript[k, e]) (Q/L)ln((((Subscript[x, 0])+L/2)/((Subscript[x, 0])-L/2)))

Solution

V=(Subscript[k, e]) ∫((dq)/r)=(Subscript[k, e]) (Q/L)(∫_(-L/2)^(L/2))(1/(Sqrt[(x^2)+(Subscript[y, 0]^2)]))dx

(Subscript[V=(Subscript[k, e]) (Q/L)ln(x+(Sqrt[(x^2)+(Subscript[y, 0]^2)]))|, -L/2]^(L/2))=(Subscript[k, e]) (Q/L)(ln(L/2+(Sqrt[(((L/2))^2)+(Subscript[y, 0]^2)]))-ln(L/2+(Sqrt[(((L/2))^2)+(Subscript[y, 0]^2)])))

V=(Subscript[k, e]) (Q/L)ln(((L/2+(Sqrt[(((L/2))^2)+(Subscript[y, 0]^2)]))/(-L/2+(Sqrt[(((L/2))^2)+(Subscript[y, 0]^2)]))))

Example . - Uniform Ring of Charge

Solution

V=(Subscript[k, e]) ∫((dq)/r)=(Subscript[k, e])∫((dq)/(Sqrt[(R^2)+(Subscript[z, 0]^2)]))=(Subscript[k, e])(1/(Sqrt[(R^2)+(Subscript[z, 0]^2)]))∫dq=(Subscript[k, e])(Q/(Sqrt[(R^2)+(Subscript[z, 0]^2)]))

Example . - Disk of Uniform Charge

Solution

V=(Subscript[k, e])∫((dq)/r)

r=(Sqrt[((r')^2)+(Subscript[z, 0]^2)])

dq=(Q/(Subscript[A, tot]))dA=(Q/(ℼ (R^2)))2π r' dr'=((2Q)/(R^2))r' dr'

V=(Subscript[k, e])∫((dq)/r)=(Subscript[k, e])((2Q)/(R^2))(∫_0^R)((r' dr')/(Sqrt[((r')^2)+(Subscript[z, 0]^2)]))

u=((r')^2)+(Subscript[z, 0]^2)  ⟹  du=2 r' dr'    and ∫(u^(-1/2))du=(1/(1/2))(u^(1/2))+C=2(u^(1/2))+C

(Subscript[V=(Subscript[k, e])((2Q)/(R^2))(Sqrt[((r')^2)+(Subscript[z, 0]^2)]) |, 0]^R)=(Subscript[k, e])((2Q)/(R^2))((Sqrt[(R^2)+(Subscript[z, 0]^2)])-(Subscript[z, 0]))

. - Potentials and Electric Fields

Potential from Electric Field

We have seen that the fundamental expression that gives the potential difference from the electric field is

ΔV=-∫ (OverVector[E])·d(OverVector[r]).

This is the change in the potential between the endpoints of the contour of the integral.  A simple special case of this is a uniform field.  If (OverVector[E]) is uniform (meaning spatially constant) then it comes out of the integral.

ΔV=-∫ (OverVector[E])·d(OverVector[r])=-(OverVector[E])·∫ d(OverVector[r])=-(OverVector[E])·Δ(OverVector[r]),

where Δ(OverVector[r]) is the vector from the starting position to the final position.

Another important special case is the infinitesimal form; if one moves along a small displacement d(OverVector[r]), the infinitesimal change in the potential is given by

dV=-(OverVector[E])·d(OverVector[r]).

Example . - Work and Potential from Field

(OverVector[E])=⟨-25,0,30⟩×(10^3)(QuantityUnit)/(QuantityUnit)

Solution

ΔV=-(OverVector[E])·Δ(OverVector[r])=-⟨-25,0,30⟩·⟨-3,2,0⟩×(10^3)(QuantityUnit)=-((-25) (-3)+0 (2)+(30)0)×(10^3)(QuantityUnit)=-75×(10^3)(QuantityUnit)

q=-e   where  e=1.60×(10^(-19))(QuantityUnit)

Δ U=q Δ V=-e Δ V=1.2×(10^(-14))(QuantityUnit)

Equipotentials

Equipotentials are surfaces of constant potential.

Potential is a scalar field, meaning that at each position in space there is a scalar function. A way of representing scalar fields in a two dimensional graph is to draw contour of constant values. On weather maps, the scalars of temperature and pressure are represented by isotherms and isobars, which are contours of constant temperature and pressure.  In three dimensions we have surfaces of constant values instead of contours. A surface of constant potential is called an equipotential and these will appear as contours in two dimensional diagrams. First consider the case of a point charge. The potential is given by: V=(Subscript[k, e])Q/r, so a surface of constant V is a surface of constant r, a sphere. The equipotential for a point charge are then concentric spheres with the charge at the center.

Field lines are perpendicular to equipotentials.

In the point charge example above, the field lines are perpendicular to the equipotentials. This is generally the case. To understand how equipotentials are related to electric field lines consider the infinitesimal form of the potential expression dV=-(OverVector[E])·d(OverVector[r]).  Take d(OverVector[r]) to be some infinitesimal displacement along the equipotential. Along the equipotential  dV=0, this is because the potential is constant along an equipotential.  We then get 0=(OverVector[E])·d(OverVector[r]) which implies that the field is perpendicular to the d(OverVector[r]). Since this is true for any direction along the equipotential it follows that the field is always perpendicular to the equipotential.

Electric field lines point toward lower potential.

Begin with the expression dV=-(OverVector[E])·d(OverVector[r]) and take d(OverVector[r]) to be some infinitesimal displacement in the direction of an electric field line.  Since the dot product of two vectors in the same direction is positive (OverVector[E])·d(OverVector[r])>0 which implies that dV=-(OverVector[E])·d(OverVector[r])<0.  This tells us that field lines always point toward lower electric potential.

Electric Field from Potential

To go from the potential to the electric field, begin with the infinitesimal expression dV=-(OverVector[E])·d(OverVector[r]).  To get (Subscript[E, x]) consider some infinitesimal displacement in the x-direction  d(OverVector[r])=(OverHat[x]) dx.

dV=-(OverVector[E])·d(OverVector[r])    and    d(OverVector[r])=(OverHat[x]) dx   ⟹   dV=-(OverVector[E])·(OverHat[x]) dx=-(Subscript[E, x])dx

Solving for (Subscript[E, x]) gives  (Subscript[E, x])=-((dV)/(dx)).  This expression is not strictly correct, because the potential may not only be a function of just x.  By moving in the x-direction we are really taking the derivative with respect to x keeping the other variables (y and z) constant.  This is what is meant by partial derivatives and the expressions for the components of the field are

(Subscript[E, x])=-((∂V)/(∂x)) ,    (Subscript[E, y])=-((∂V)/(∂y)) ,     (Subscript[E, z])=-((∂V)/(∂z)) .

The radial part of the field can also be found in a similar manner and for problems with spherical symmetry, the above partial derivative becomes the ordinary derivative

(Subscript[E, r])=-((∂V)/(∂r))    and   (Subscript[E, r])=-((dV)/(dr))  (for spherical symmetry)

It should be mentioned as an aside that the above relations can be written as a single vector expression using the gradient operator of vector calculus.

(OverVector[E])=-(OverVector[∇])V

Example . - Partial Derivatives and Electric Fields from Potential

f(x,y)=a (x^3)y-b (y^2)

Solution

((∂f)/(∂x))=3a (x^2)y   and   ((∂f)/(∂y))=a (x^3)-2b y

V(x,y)=a (x^3)y-b (y^2)

Solution

(Subscript[E, x])=-((∂V)/(∂x))=-3a (x^2)y   and   (Subscript[E, y])=-((∂V)/(∂y))=-a (x^3)+2b y

(OverVector[E])=⟨(Subscript[E, x]),(Subscript[E, y])⟩=⟨-3a (x^2)y ,-a (x^3)+2b y⟩

Conductors in Electrostatics - II

Potential is constant throughout a conductor in electrostatics.

The condition that the electric field is zero implies that the potential is constant.  Consider some contour entirely inside a conductor.  Integrating the field to get the potential difference gives

ΔV=-∫ (OverVector[E])·d(OverVector[r])=0.

Note that this argument does not apply to disconnected conductors.  If there are several disconnected conductors each one will have its own potential.

It follows that the surface of a conductor is always an equipotential.  The condition that the field is perpendicular to a conductor may now be seen as a consequence of fields being perpendicular to equipotentials.

Example . - Field Lines, Equipotentials and Conductors

Solution

Solution

Solution

Example . - A Point Charge Inside a Hollow Conducting Sphere (Continued)

Solution

Forr<a:    (Subscript[E, r])=(Subscript[k, e])(q/(r^2))

Fora<r<b:    (Subscript[E, r])=0

Forr>b:   (Subscript[E, r])=(Subscript[k, e])((q+Q)/(r^2))

(Subscript[E, r])=-((dV)/(dr)) .

For r>b: ((dV)/(dr))=-(Subscript[E, r])=-(Subscript[k, e])((q+Q)/(r^2))   ⟹  V(r)=(Subscript[k, e])((q+Q)/r)+(Subscript[C, 1])

V(∞)=0  ⟹  0+(Subscript[C, 1])=0  ⟹  V(r)=(Subscript[k, e])((q+Q)/r)

(lim_(  r->(SuperMinus[b])))V(r)=(lim_(  r->(SuperPlus[b])))V(r)   ⟺  V((SuperMinus[b]))=V((SuperPlus[b]))

For a<r<b: ((dV)/(dr))=-(Subscript[E, r])=0   ⟹  V(r)=(Subscript[C, 2])

V((SuperMinus[b]))=V((SuperPlus[b]))  ⟹  (Subscript[C, 2])=(Subscript[k, e])((q+Q)/b)  ⟹  V(r)=(Subscript[k, e])((q+Q)/b)

For r<a: ((dV)/(dr))=-(Subscript[E, r])=-(Subscript[k, e])(q/(r^2))   ⟹  V(r)=(Subscript[k, e])(q/r)+(Subscript[C, 3])

V((SuperMinus[a]))=V((SuperPlus[a]))   ⟹  (Subscript[k, e])(q/a)+(Subscript[C, 3])=(Subscript[k, e])((q+Q)/b)  ⟹  V(r)=(Subscript[k, e])(q/r)+(Subscript[k, e])((q+Q)/b)-(Subscript[k, e])(q/a)