Conductors in Electrostatic Equilibrium

AP Physics C

Montwood High School

R. Casao

• A good electrical conductor contains electrons that are not bound to any atom and are free to move within the conductor.

• If the electric field E is not zero in some area, the electrons there feel a force F = E·q and start to move.

• The electrons adjust their positions until the force on every electron is zero.

• When there is no net motion of charges within a conductor, the conductor is in electrostatic equilibrium.

• Properties of a conductor in electrostatic equilibrium:

– The electric field is zero everywhere inside the material.

– Any excess charge on an isolated conductor must be entirely on its surface.

– The electric field just outside a charged conductor is perpendicular to the conductor’s surface and has magnitude 2··k· (or /o).

– On an irregularly shaped conductor, charge tends to accumulate at locations where the radius of curvature is the smallest (at sharp points).

Under electrostatic conditions, any excess charge resides entirely on the surface of a

solid conductor.

Gaussian

surface is

labeled A

• Generally draw a Gaussian surface to satisfy two conditions: –Condition 1: electric field constant

over the surface by symmetry.–Condition 2: the dot product of

EdA = E·A·cos because E and A are parallel ( = 0°).

• Proof that electric field just outside a charged conductor is 2··k· (or /o).

• Consider a section of the surface small enough to neglect any curvature and take the section to be flat.

• Imagine a tiny cylindrical Gaussian surface to be embedded in the section. One end is fully inside the conductor, the other

end is fully outside the conductor and the cylinder is perpendicular to the conductor’s surface.

• E at and just outside the conductor’s surface must be perpendicular to the surface. If not, a component of E would exist in the direction of the conductor’s surface that would exert forces on the surface charges and cause them to move.

• There is no flux thru the internal end of the cylinder because no electric field lines pass through the inner surface of the Gaussian cylinder because the electric field E inside the conductor is zero.

• There is no flux thru the curved surface of the cylinder because E = 0 N/C within the

conductor and externally E is parallel to the curved portion of the Gaussian cylinder.

• The only flux thru the Gaussian cylinder occurs thru the external end of the cylinder where E is perpendicular to A. The charge enclosed

(qin) by the Gaussian surface lies on the conductor’s surface in an area A; = Q/A

so qin = ·A.

• The magnitude of E just outside a conductor is proportional to the surface charge density at that location on the conductor.

• If the charge on the conductor is positive, E is directed away from the conductor.

• If the charge on the conductor is negative, E is directed toward the conductor.

Spherical Gaussian surfaces around (a) positive and (b) negative point charge.

o

in

in

in

in

ε

σσkπ4E

σkπ4Lrπ2

σLrπ2kπ4

A

qkπ4E

σLrπ2A

QLrπ2q1θcos

θcosAEqkπ4ΦdAEΦ

Lrπ2

q

A

Q:iprelationshalproportion

A

Qσ

Off-Centered Charge Distribution

• Consider a –5 C point charge located closer to one side of an electrically neutral shell.

• A charge of +5 C will lie on the inner wall of the shell; the E inside the shell must be 0 N/C; qin = -5 C + 5 C = 0 C.

• If the point charge were centered inside the shell, the positive charge would be uniformly distributed along the inner wall. Since the point charge is off-center, the distribution of positive charge tends to collect on the section of the wall nearest the -5 C.

• Because the shell is electrically neutral, its inner wall can only have a charge of +5 C if electrons with a total charge of -5 C leave the inner wall and move to the outer wall.

• There they spread out uniformly because the shell is spherical and because the distribution of positive charge on the inner wall cannot produce an electric field in the shell to affect the distribution of charge on the outer wall.

Example 24.7 A

Sphere Inside a

Spherical Shell

Example 24.7 A Sphere Inside a Spherical Shell

• Consider a solid conducting sphere of radius a surrounded by a conducting spherical shell of inner radius b and outer radius c.

• The net charge on the solid conducting sphere is +2·Q. The electric field E from the solid conducting sphere radiates outward from the sphere of radius a and would be uniform.

• The net charge on the conducting spherical shell is –Q. The electric field E from the conducting spherical shell radiates inward and would be uniform.

• The entire system is in electrostatic equilibrium.

• The spherical nature of the objects satisfies condition 1 (electric field E is constant over the surface by symmetry).

• Region 1: the charge 2·Q will lie on the outer surface of the solid conducting sphere of radius a.

• Draw a spherical Gaussian surface of radius r < a.

• No electric field E can exist within the conductor. – The negative charges from the outer

conducting spherical shell attracts the positive charge on the solid conducting sphere to the surface.

– The repulsion of the positive charges on the solid conducting sphere also contributes to the positioning of the positive charge on the surface of the solid conducting sphere.

– No charge lies within the solid sphere, therefore, qin = 0 C.

• Region 2: the charge 2·Q will lie on the outer surface of the solid conducting sphere of radius a.

• Draw a spherical Gaussian surface of radius r a.

N/C0E0qr

qk

rπ4

qkπ4

A

qkπ4E

rπ4A1θcos

θcosAEqkπ4qkπ4Φ

θcosAEdAEdAEΦ

in

2in

2inin

2

inin

• Charge located inside the spherical Gaussian surface of radius r is 2·Q; therefore, qin = 2·Q

• Spherical Gaussian surface satisfies condition 1: electric field constant over the surface by symmetry.

• Spherical Gaussian surface satisfies condition 2: the dot product of EdA = E·A·cos because E and A are parallel ( = 0°).

22in

2in

2inin

in

2in

r

Qk2E

r

Q2kEQ2q

r

qk

rπ4

qkπ4

θcosA

qkπ4E

θcosAEqkπ4

rπ4A1θcosqkπ4Φ

θcosAEdAEdAEΦ

• Region 3: inside the conducting spherical shell which is in electrostatic equilibrium:– The negative charges on this conductor are

located on the inner surface of the conducting spherical shell because they are attracted to the positive charge located on the solid conducting sphere.

– The charge on the inner surface of the spherical shell must be -2·Q to balance the +2·Q charge located on the solid sphere. Therefore, a charge of +Q must reside on the outer surface of the conducting shell, bringing the net charge to –Q; (-2·Q +Q = -Q).

– Qnet = Qinner surface + Qouter surface

– The distribution of negative charge on the inner surface and the positive charge on the outer surface of the shell means that there is no charge located within the spherical shell itself because all the charge is located on the surface.

– Draw a spherical Gaussian surface of radius b r < c.

– The spherical Gaussian surface satisfies condition 1 and condition 2; the electric field E can be considered to be constant over the surface by symmetry and the dot product of EdA = E·A·cos because E and A are parallel ( = 0°).

N/C0E0qr

qk

rπ4

qkπ4

A

qkπ4E

rπ4A1θcos

θcosAEqkπ4qkπ4Φ

θcosAEdAEdAEΦ

in

2in

2inin

2

inin

• Region 4: lies outside the conducting spherical shell.– A spherical Gaussian surface can be drawn to

enclose the conducting spherical shell and the solid conducting sphere.

– The radius of the spherical Gaussian surface is r c.

– The spherical Gaussian surface satisfies condition 1 and condition 2; the electric field E can be considered to be constant over the surface by symmetry and the dot product of EdA = E·A·cos because E and A are parallel ( = 0°).

chargepointawereitifasr

QkE

QQQ2Q2qr

qk

rπ4

qkπ4

A

qkπ4E

rπ4A1θcos

θcosAEqkπ4qkπ4Φ

θcosAEdAEdAEΦ

2

in

2in

2inin

2

inin