Conductors

4.5 Materials in the Electric Field

4.5.1 Conductors

the negative and positive charges in the conductor and the charges move to the surface. Because these charges were not introduced externally (i.e., the conductor has not acquired additional charge), they are calledinduced charges. When the force that has induced the charges is removed, the charges return to their neutral state.

Properties of conductors in the electrostatic field are as follows:

(1) Charge introduced into a conductor moves to the surface of the conductor. No charge exists in the interior.

(2) Charges distribute themselves on the surface of the conductor to produce zero electric field intensity in the interior of the conductor, regardless of the shape of the conductor or the external charges that may exist. This property is one of the most useful properties in application of conductors in electrostatics.

(3) A single point charge may exist anywhere in the volume of a conductor because there are no forces acting on it. In practice, this cannot happen unless the charge is a single electron. Any other charge will redistribute itself on the surface of the conductor.

(4) The volume charge density inside a conductor is zero. Surface charges may exist on conducting surfaces.

(5) The electric field intensity on the surface of the conductor (external to the surface) must be perpendicular to the surface at any point on the surface. Because of this, the potential on the surface of any conductor is constant.

The latter property is explained using Figure 4.22as follows. If a charge exists on the surface of a conductor and it produces both a normal and a tangential electric field, the normal component“pushes”any other charge against the surface.

Since the charge cannot escape the conductor, this component of the electric field cannot move the charge. The tangential component will generate a force (with other charges in the vicinity), which will force the charges to move along the surface.

This motion can only stop if there is no force to sustain it (remember, charges can move freely in the conductor). The charges in a conductor will come to rest in a configuration which cancels the tangential (parallel to the surface) electric field component. The only remaining force (and therefore electric field) is perpendicular to the surface.

4.5.1.1 Electric Field at the Surface of a Conductor

We mentioned that the tangential component of the electric field intensity at the surface of a conductor must be zero simply from the definition of the conductor. What can we say about the normal component? To consider this, we useFigure4.23.

A charge density is distributed on the surface of an arbitrary conductor. As required, we assume that only a normal component of the electric field intensity exists and the electric field intensity inside the conductor must be zero. By use of Gauss’s law, we can write

q q

F F

a b

q

q F

F

E=0 F_tang F_normal

F_normal F_tang

+

+

+

+

Figure 4.22 The process by which tangential components of the electric field intensity at the surface of a conductor vanish. (a) Tangential components force the charges to move.

(b) Charges stop moving when the tangential field components cancel

+ + + + + + + + + +

+ + + + + + + + + +

− − − − − − − − − −

− − − − − − − − − −

conductor conductor

a

b Figure 4.21 Charging of a

conductor by induction. (a) Conductor in neutral state.

(b) Separation of charges by an external electric field

En¼ρs

ε0

V

m ð4:36Þ

Thus, the conditions at the surface of a conductor are given as Et¼0, En¼ρs

ε0

V

m , or Dn¼ρs

C m²

ð4:37Þ

The relation inEq. (4.37)⁴not only gives the electric flux density on the surface but also allows the calculation of the charge density on the surface of the conductor when the conductor is inserted into an electric field. We will make considerable use of this property in examples and in many instances in future chapters.

Example 4.13Surface Charge Density

The electric field intensity at the surface of a conductor is known to be perpendicular, pointing outward, equal to 100 V/m, and the conductor is in free space:

(a) What is the charge density on the surface of the conductor?

(b) What are the electric field intensity and the electric potential inside the conductor if the potential on the surface of the conductor isVs ¼10 V?

Solution: The charge density at the surface of a conductor is found usingEq. (4.36)or(4.37). The electric field intensity in any conductor under static conditions is zero; the potential is constant and equal to the surface potential.

Thus, (a)

ρs¼Dn ¼ε0En¼8:85410¹²100¼8:85410¹⁰ C m²

The charge density is positive since the electric field intensity points outward.

(b)

E¼0 V

m , V¼Vs¼10 ½ V

Example 4.14Conductor in an Electrostatic Field

A point chargeQ[C] is surrounded by a hollow conducting sphere as shown inFigure4.24a. Calculate:

(a) The electric field intensity and electric potential everywhere in space. Plot the electric field intensity and the potential.

(b) The surface charge densities on the inner and outer surfaces of the conducting shell.

E E

n

ρs

+

+ + +

+ +

+ conductor air

Figure 4.23 Charge density and electric field intensity at the surface of a conductor

4These properties, namely, that the electric field intensity at the surface of a conductor is normal to the surface and directly proportional to the surface charge density, were discovered by Coulomb in the course of his investigations on charged bodies.

168 4 Gauss’s Law and the Electric Potential

Solution: The electric field intensity inside the conducting sphere is always zero, but everywhere else, the field of the point charge is not disturbed. This means that the inner surface of the spherical shell has a total charge equal to –Quniformly distributed on the surface and the outer surface acquires a total positive chargeQ. These can be found from direct application of Gauss’s law. After the electric field intensity is found everywhere, the electric potential is found by integrating from1to the required point:

(a) A Gaussian surface atR<ais used to find the electric field intensity inside the shell and one atR>bto find the electric field intensity outside the sphere: these are shown inFigure4.24b. Since the electric field of a point charge is radial (and, therefore, perpendicular to any spherical surface), we have from Gauss’s law forR1andR3,

ð

s

E

^{ds¼E4πR21¼}_ε^Q

0

! E¼ Q 4πR²1ε0

V

m , 0R2<a E¼0, aR2b

ð

s

E

^ds¼E4πR23 ¼Q ε0

! E¼ Q 4πR²3ε0

V

m , R3 >b

The electric potential is found by integrating from1to the required point. Outside the sphere,R ¼R3>b:

VR₃¼ ðR3

1

E

^{dl¼ ðR3}

1

R^E

^{R^dR¼ ðR3}

1

QdR

4πε0R²¼ Q 4πε0R3

½ V

At the outer surface of the conductor, the potential is found by settingR3¼b:

Vb¼ Q 4πε0b ½ V

To find the potential inside the conductor, we integrate fromR¼btoR ¼R2overE. Using the same relation as forVR₂

we get

VR₂ ¼ ðR2

1E

^{dl¼ ðb}

1

QdR 4πε0R²

ðR2

b

0dR¼ Q

4πε0b ½ V

Thus, the electric potential remains constant in the conductor. We know this to be true from the fact that a perfect conductor is always at constant potential. The potential inside the sphere (0<R1<a) is

VR₁¼ ðR1

1

E

^{dl¼ ðb}

1

QdR 4πε0R²

ða

b

0dR ðR1

a

QdR 4πε0R²¼ Q

4πε0

1 b1

aþ 1 R1

½ V

a b

Q E=0

a b

E

R

a

a b c

b R

V

a b

Q

s1

s₂ s₃ R₁

R₂ R₃

−ρ^a +ρb

Q/4πε0R12

Q/4πε0R32

Q/4πε0R3

Q/4πε^0b

Q(1/b−1/a+1/R1)/4πε0

conductor

Figure 4.24 Electric field due to a point charge in the presence of a conductor. (a) Point charge surrounded by a hollow conducting sphere. (b) The Gaussian surfaces used to calculate the electric field intensity. (c) Plot of the electric field intensity and potential

The potential atR1¼a equals that at band the potential at the location of the point charge (R1¼0) is infinite as required. A plot of the electric field intensity and electric potential is shown inFigure4.24c. Note that the integration is done in segments because the expression for the electric field intensity in each segment is different.

(b) To calculate the charge densities on the inner surface, we use the fact thatE ¼0 inside the conductor. We draw a Gaussian surface atR¼a⁺(immediately inside the conductor so that the surface is included). For this Gaussian surface to produce zero field intensity, the total charge enclosed must be zero:

ð

s

E

^ds¼E4πR22 ¼Q ε0

þρsa4πa² ε0

¼0 ! ρsa¼ Q 4πa²

C m²

The surface charge density on the outer surface is calculated from the fact that the total charge in the system isQ[C].

Therefore, the total charge on the outer surface equals the total charge on the inner surface. This gives Q_b¼ Q_a¼Q ! ρsb¼ Q

4πb² C m²