Electric Potential Due to Distributed Charges

4.4 The Electric Potential

4.4.2 Electric Potential Due to Distributed Charges

This simple calculation indicates the ease with which the potential due to any number of point charges may be calculated.

Note also that in the final result, the term in the square brackets is a geometrical term and does not include charge or material properties. This will be exploited inChapter6to find numerical solutions to a number of much more complex problems.

Exercise 4.3 Three charges, each equal toQ¼10^–7C, are located on thexaxis atx¼0,x¼1 m, andx¼2 m, as shown inFigure4.13. Calculate the potential at (x¼0, y¼1).

Answer 1,936 V.

Thus, the net effect is replacement of the elemental charge inEq. (4.30)by the infinitesimal charge and replacement of the summation by integration. The charge densities were assumed to be location dependent and, therefore, left inside the integral sign. If charge density is uniform, it should be taken out of the integral. Also, note that the integration is performed over the primed coordinates; that is, only on that part of space over which the charge density is nonzero.

InEqs. (4.31)through(4.33), we have used the potential of an infinitesimal (point) charge to calculate the potential of a charge distribution. The main premise is that the scalar summation (integration) of the potentials of point charges can be performed relatively easily. This is certainly true in some cases but not always. Sometimes, in particular when highly symmetric charge distributions are involved, it is simpler to calculate the electric field intensity first, using Gauss’s law, and then use the definition of potential inEq. (4.20)to evaluate the potential. The following examples explore both methods.

Example 4.8 A short segment of lengthL ¼2 m is charged with a uniform line charge densityρl¼10^–7C/m:

(a) Calculate the potential at a general point in space.

(b) What is the potential at a radial distanced¼0.1 m from the center of the segment?

Solution: The geometry is shown inFigure4.15where the segment was placed in a cylindrical system of coordinates.

Clearly, Gauss’s law cannot be used to calculateE.Thus, an element of lengthdz⁰is chosen at location z⁰, with charge dq¼ρl dz⁰. The distance between this point and the general point P(r,ϕ,z) is calculated and the potential found by integration along the segment:

(a) Although, in general, the distance between pointz⁰andPdepends on all three coordinates, in this case it is independent of the angleϕ. The distance |r| is

r¼ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²þðzz⁰Þ² q

½ m P(x,y,z)

V(R) Δl^'

P(x,y,z) V(R) ρ^s

P(x,y,z) V(R) Δv^'

ρ^v

x y

z

R R R

x y

z x

y

z

R^i' R^i' R^i'

|R−R'i| |R−R'i| |R−R'i|

ΔQi=Δviρ^v ΔQi=Δsiρ^s

Δs'

ΔQi=Δl^i'ρ^l s' ^{' v' '}

a b c

Figure 4.14 Calculation of potential due to (a) line charge distribution, (b) surface charge distribution, (c) volume charge distribution

ρl φ z'

z

r

z'=−L/2 z'=L/2

x

dl'=zdz'

P(r,φ,z)

P(r,φ,0) P(0.1,φ,0) z'

d=0.1m

z−z'

Figure 4.15 Calculation of potential atP(r,ϕ,z) due to a short, charged line segment

158 4 Gauss’s Law and the Electric Potential

The potential due to the element of charge (with respect to zero potential at infinity) is [seeEq.(4.30)]

dV¼ ρldz⁰ 4πε0

r¼ ρldz⁰ 4πε0

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²þðzz⁰Þ²

q ½ V

Integrating this fromz⁰¼–L/2 toz⁰ ¼+L/2 gives V r;ð ϕ;zÞ ¼ ρl

4πε0

ðz⁰¼L=2

z⁰¼L=2

dz⁰ ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²þðzz⁰Þ²

q ¼ ρl

4πε0

ln 2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²þðzz⁰Þ² q

þ2z⁰2z

L=2 L=2

¼ ρl

4πε0

ln 2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²þðzL=2Þ²

q þL2z

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²þðzL=2Þ²

q L2z

0 B@

1 CA ½ V

ForL¼2 m,ρl¼10^–7C/m, andε0¼8.85410^–12F/m, the potential at a general point is V r;ð ϕ;zÞ ¼ 10⁷

4π8:85410¹²ln 2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²þðz1Þ²

q þ22z

2

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²þðzþ1Þ²

q 22z

0 B@

1

CA¼898:8 ln

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²þðz1Þ²

q þ1z

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi r²þðzþ1Þ²

q 1z

0 B@

1 CA ½ V

(b) At a radial distanced ¼0.1 m (pointP(0.1,ϕ,0)) the potential is

Vð0:1;ϕ;0Þ ¼898:8 ln

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:1²þ 1ð Þ²

q þ1

ffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffiffi 0:1²þð Þ1 ²

q 1

0 B@

1

CA¼5389:6 ½ V

Example 4.9Potential Due to a Charge Distribution A very long conducting cylinder of radiusa[m] is coated with a dielectric layer and connected to zero potential as shown inFigure4.16a. A uniform volume charge density ρv¼ρ0[C/m³] is distributed throughout the dielectric layer:

(a) Calculate the potential everywhere inside the charge layer.

(b) Calculate the potential outside the charge layer.

(c) Plot the potential in space.

Solution: It is simpler to calculate the electric field intensity first, using Gauss’s law, because the structure is very long.

After the electric field is calculated everywhere, the potential is evaluated by integration, usingEq. (4.20). The Gaussian surfaces needed for the various regions are shown inFigure4.16b.

r r

ε r

r=a r=b ε⁰

ε⁰ ρ=ρ⁰

ε⁰ ρ=ρ⁰

2

ε ε⁰

s₀

s₁ s

v

v

dielectric material Figure 4.16 Calculation of

potential in a cylindrical geometry shown in cross section. (a) Configuration.

(b) Gaussian surfaces needed to calculate the electric field intensity

(a)Electric field intensity inside the inner cylinder:r<a.

E¼0 (no charge inside any Gaussian surface drawn in this region).

Between the two cylinders:ar<b.

The Gaussian surface is a cylinder of radiusr(r>a) and arbitrary lengthL:

ð

s₁

E

^ds1¼1_ε^ð

v⁰₁

ρvdv⁰₁

wherev1⁰ is that part of the volume enclosed by the Gaussian surfaces1that contains charge density. For a cylindrical Gaussian surface atr(a<r<b), the volume that contains charge is a hollow cylinder of lengthL, inner radiusa, and outer radiusr. Because charge density is uniform, the integral on the right-hand side is

ð

v⁰₁

ρvdv⁰₁¼ρ0πr²πa² L

Substituting in Gauss’s law, and since bothdsandEpoint radially out, E2πrL¼ρ0ðπr²πa²ÞL

ε ! E¼^rρ0ðr²a²Þ 2rε

V m

Note: In the dielectric, we use εnotε0! The potential inside the charge layer may be calculated from the definition of potential inEq. (4.20):

Vr₀,a ¼Vr₀Va¼Vr₀¼ ðr0

a

E

^{dl¼ ðr0}

a

^

rρ0ðr²a²Þ

2rε

^{dl ½ V}

The integration is along the radial direction fromato an arbitrary pointa <r0<banddl¼^rdr:

Vr0¼ ðr₀ a

^

rρ0ðr²a²Þ

2rε

^{r^dr¼ ρ}₂⁰_ε^ðr0

a

r²a² ð Þ

r dr ¼ ρ0

2ε r²

2 a²lnr 2

4

3

5

j

^r_a^{0¼ ρ4ε024r20a22a2lnra035 ½ V}

Note: The constant of integration is zero since the reference voltageVa¼0. Thus, atr0¼b, the potential is Vb¼ ρ

4ε b²a²2a²lnb a

½ V

(b) For the domain outside the outer cylinder,b<r<1, again using a Gaussian surface ats2, we get E¼^rρ0b²a²

2rε0

V m

This is the same form as for the domain inside the dielectric, but the total charge is contained betweenr¼aand r¼b.

To calculate the potential at a pointr1outside, we integrate as in the previous case fromr¼btor¼r1: Vr₁,b¼Vr₁Vb¼

ðr1

b

ρ0b²a² 2rε0

dr¼ ρ0b²a² 2ε0

lnr1

b ! Vr₁¼Vbρ0b²a² 2ε0

lnr1

b ½ V

160 4 Gauss’s Law and the Electric Potential

In this case, the reference point is atr¼bandVbwas calculated above. Substituting this, we get Vr₁,b¼ ρ0

4ε b²a²2a²lnb a

ρ0b²a² 2ε0

lnr1

b ½ V

(c) See Figure 4.17 for a sketch of the potential. Note that the potential is continuous everywhere. The potential at r¼b ¼0.2 m is113.91 V.

Example 4.10 Two very long, thick cylindrical conducting lines, each of radiusa[m], are separated a distanced [m]. A charge densityρs[C/m²] is distributed uniformly on the surface of one line and –ρs[C/m²] on the surface of the second, as shown inFigure4.18a:

(a) Calculate the potential difference between the two lines.

(b) Can you tell what the absolute potential of each line is?

Solution: The electric field intensity everywhere in space may be calculated from Gauss’s law. To calculate the potential of the lines, it is necessary that we find a reference point at which the potential is known. The solution at infinity cannot be used as a reference potential because the lines themselves extend to infinity, but the point midway between the two lines is at a constant potential anywhere along the line. In fact, the potential is zero although the actual value is not important. Another possibility is to assume one line is at a constant (perhaps zero) potential and calculate the potential of the other relative to the first:

(a) Suppose we choose a general pointP(x,0) and calculate the electric field intensity at this point usingFigure4.18b. The electric field intensity due to lineAis found by constructing a Gaussian surface around conductorAin the form of a cylinder of radiusrAand lengthL. From Gauss’s law, we get

EA2πrAL¼ρs2πaL ε0

! EA¼ ρsa ε0rA

V m

The solution for conductorBis identical except that the electric field intensity is negative (pointing toward the conductor):

EB2πrBL¼ ρs2πaL ε0

! EB¼ ρsa ε0rB

V m a

A B

a b

d

d/2

P a x

a a

A B

d/2 x +ρs

+ρ^s −ρ^s −ρs

EA

EB

rA

rB

Figure 4.18 Two thick lines with uniform surface charge density. (a) Geometry and dimensions. (b) Calculation of electric field intensity at pointP

0.1 0.2 0.3 0.4

−1200

−800

−400

0 r

−1288

−113.9

distance [m]

voltage [V]

Figure 4.17 The potential everywhere in space due to the configuration inFigure4.16a

whereρs2πaLis the total charge on a cylinder of radiusaand lengthL. In vector notation, EA ¼r^A

ρsa ε0rA

, EB¼r^B

ρsa ε0rB

V m

From geometrical considerations alone, the potential midway between the two conductors must be zero whereas the electric field intensity on the line connecting their centers is horizontal. However, we will assume the potential midway is some fixed valueV0for generality. UsingFigure4.18band taking pointPat a general location between the cylinders, we can write

rA ¼d

2þx, rB¼d

2x ½ m The electric field intensity is in the positivexdirection and equals

Eðx;0Þ ¼x^ ρsa ε0rA

þ ρsa ε0rB

¼x^ρsa ε0

2

dþ2xþ 2 d2x

V m

To find the potential on lineB, we integrate in the direction of the electric field fromx¼0 tox¼(d/2) –a. In this case, dl¼x^dx:

VB,0 ¼VBV0¼ ðd=2a

x¼:

E

^{dl¼ ðd=2a}

x¼0

^ x ρsa

ε0

2

dþ2xþ 2 d2x 0

@

1 A

^{x^dx}

¼ 2ρsa ε0

ðd=2a

x¼0

1

dþ2xþ 1 d2x 0

@

1

A¼ ρsa ε0

lndþ2x d2x

d=2a

0

¼ ρsa ε0

lnda

a ½ V

To obtain the potential on line A, we integrate against the electric field from x¼0 to x¼–(d/2) +a, again with dl¼x^dx. The result is a positive potential differenceVA0of identical magnitude:

VA,0¼VAV0¼ ðd=2þA

x¼0 E

^{dl¼ 2ρ}_ε^sa

0

ðd=2þa

x¼0

1

dþ2xþ 1 d2x 0

@

1 A¼ ρsa

ε0

lnda

a ½ V

The potential difference between linesAandBis

VAB¼VA,0VB,0 ¼VAVB¼ρsa ε0

lnda d

ρsa ε0

lnda d 0

@

1 A¼2ρsa

ε0

lnda a ½ V Note thatV0cancels, meaning that the reference potential is irrelevant.

(b) In general, we cannot calculate the absolute potential of the lines. However, in this case, the lines are charged with equal and opposite charge densities. This means that the potential midway between them is zero and the potentials calculated above are the absolute potentials. If, on the other hand, the potential at the center between the lines is raised by, say,V0, then the potentialsVA andVBare also raised by the same value. Under this condition, only the potential difference between the lines is known.

162 4 Gauss’s Law and the Electric Potential

Exercise 4.4 Two parallel plates are very large in extent (infinite), separated a distanced[m], in air, and charged with equal but opposite charge densityρs[C/m²]. Calculate the potential difference between the two plates.

Answer V ¼ρsd ε0

½ .V

Exercise 4.5 Two parallel plates are very large in extent (infinite), separated a distance d and connected to a potential differenceV. Calculate the electric field intensity between the two plates.

Answer E¼V

d [V/m], pointing from the positive to the negative plate.