physics 227: lecture 5 gaussâ€™s law leftovers, work

Physics 227: Lecture 5Gauss’s Law Leftovers, Work /

Potential Energy in Electrostatics• Lecture 4 review:

• Gauss’s Law: for a closed surface, Φ = ∫ E.dA = ∫ E⊥dA = ∫ E cosφ dA = qenclosed/ε0.

• The flux does not depend on the size or shape of the Gaussian surface, or the position of the charges, it depends only on the total enclosed charge.

• Examples: point charge, inside uniform sphere, inside spherical shell, infinitely long line charge, infinitely long cylinder of charge, inside the cylinder of charge, infinite plane of charge, two planes of charge, ...

• Conductors: electrons move around so that the field inside a conductor vanishes. Thus requires all net charge is on the surface.

Monday, September 19, 2011

Some Geometry

• The area corresponding to an angle range dθ x dφ is r2 dθ dφ, if the surface is perpendicular to r vector.

• If the surface is tilted by an angle α - cos α = r.n - its actual area in the range of angles dθ x dφ increases by a factor 1/cosα.

• With charge density λ, the total charge on the surface is: dq = rdθ rdφ λ / cosα.

rdφ dθ

r dφ

r dθ

n

ˆˆ


Field Inside Spherical Shell• For an spherical or infinite cylindrical shell of charge, the field inside is 0.

• Here we show this occurs because the fields from opposite parts of the sphere or the shell cancel.

• What is the field at the diamond a distance z above the center of the circle - a 2d slice through a cylinder or sphere.

• To get the field, we need to integrate over angles: ``E = ∫kdq/r2’’. The geometry is shown for a particular angle θ. (Angles are measured from (0,z), not the origin.)• The field from a small area at the upper right, a distance r away from (0,z), will be kdq/r2 = k (rdθ rdφ λ / cosα) /r2 = k (dθ dφ λ / cosα). This is independent of r! And independent of α for opposite points from (0,z)!

• The field from the opposite direction is exactly the same magnitude, so the two cancel! This only works for the circular geometry - for, e.g., a cube the factors of cosα would generally be different.

r0

r0

α

α

z

θ


Field at Surface of a Charged Conductor

• What is the field at the surface of a charged conductor?• Draw a small Gaussian box across the surface. The box should be small enough that the surface is flat and the charge density is constant, to a good approximation, and should be || / ⊥ to the local surface.

• Gauss’s Law: Φ = ∫E.dA = 4.0.A + 0.A + E.A = EA• Φ = q/ε0 = σA/ε0• ➮ E = σ/ε0• This is a factor of 2 larger than for a plane of charge, because all the field is to one side.


E Field in a Idealized Parallel Plate Capacitor

• (Superposition) Adding the fields from two infinite planes of charge yields the idealized result.

• Or we can use Gaussian cylinders + properties of conductors + symmetry: E = E0ŷ, but E = 0 at the end of the Gauusian cylinder inside the conductor, and E⊥ = 0 for sides of the Gaussian cylinders.

• For S2 and S3, qenclosed = 0 ➮ Φ = 0 ➮ Eoutside.A = 0 ➮ E = 0.

• For S1, qenclosed = σA ➮ Φ = σA/ε0 ➮ Einside.A = σA/ε0 ➮ Einside = σ/ε0 ŷ.• For S4, qenclosed = -σA ➮ Φ = -σA/ε0 ➮ Einside.A = -σA/ε0 ➮ Einside = σ/ε0 ŷ. (Note: here the normal is in the -ŷ direction, so the negative signs cancel.)

• Do not get confused - we have ``3’’ ways of getting the same result. Do not add them! Even though there is ±σ, you can get the field using one of them! The other is already included in your work in using Eoutside, in conductor = 0.


iClicker: What is the Surface Charge on a Conductor

Consider a conducting sphere, with a cavity inside.We insert a 5 μC charge into the cavity.What is the charge on the inner and outer surfaces of the conductor?

A. 0 μC, 0 μC.

B. -5 μC, -5 μC.

C. -5 μC, 5 μC.

D. 5 μC, -5 μC.

E. 5 μC, 5 μC.

5 μC


iClicker: What is the Surface Charge on a Conductor

Consider a conducting sphere, with a cavity inside.We insert a 5 μC charge into the cavity.What is the charge on the inner and outer surfaces of the conductor?

A. 0 μC, 0 μC.

B. -5 μC, -5 μC.

C. -5 μC, 5 μC.

D. 5 μC, -5 μC.

E. 5 μC, 5 μC.

5 μC

We need -5 μC on the inside surface to stop the field lines from the +5 μC charge in the cavity. With a neutral conductor, that requires we have +5 μC on the outer surface. If the conductor had total charge Q μC, we would have Q+5 μC on the outer surface.


Conservative Forces

• Gravity and Electric Forces are conservative forces.

• You can relate the force to a potential / potential energy.

• Energy is converted from potential to kinetic. No energy is lost.

• Changes in the potential / potential energy, depend only on the initial and final points of a trajectory, not on the intermediate path taken.


Force / Work / Potential Energy / Field / Potential

• Let’s review these concepts from gravity first - for a mass m in a constant gravitational field, like near the surface of the earth

W =�

�F · d�z = −mg∆z

∆U = −�

�F · d�z = mg∆z

Work done by gravity

Force of gravity

Potential energy from gravity often used as

Gravitational field

�g = �Fg/m = −gz

�Fg = −mgz

U = mgz

Just as the field = the force / m, the potential = the potential energy / m

integrate

differentiate

divide by m

divide by m

integrate

differentiate

∆V = −�

�g · d�z = g∆z



Work done by electric force

Electric Force

Potential energy from electric force often used as

Electric field Just as the field = the force / q, the potential = the potential energy / q

integrate

differentiate

divide by q

divide by q

integrate

differentiate

• Now we move to a charge q in a constant electric field, as with an infinite plane of charge.

U = qEz

∆U = −�

�Fc · d�z = qE∆z

W =�

�Fc · d�z = −qE∆z

∆V = −�

�E · d�z = E∆z

�Fc = q �E = −qσ

2�0z

�E = �F/q = − σ

2�0z



Work in units of Joules

Electric Force in units of Newtons

Energy in units of Joules (or “electron Volts”)

Electric field units: Volts/m or Newtons/Coulomb

Potential V in units of Volts = J/C

integrate

differentiate

divide by q

divide by q

integrate

differentiate

• Now we move to a charge q in a constant electric field, as with an infinite plane of charge.

U = qEz

∆U = −�

�Fc · d�z = qE∆z

W =�

�Fc · d�z = −qE∆z

∆V = −�

�E · d�z = E∆z�E = �F/q = − σ

2�0z

�Fc = q �E = −qσ

2�0z


iClicker: Work and Potential energy

Note: in all those formulas you have to watch your limits of integration / signs. Let’s see about your intuition!The field points left.The 5 μC charge moves right.Which of the following is correct?

A. The field does positive work on the charge and potential energy increases.

B. The field does positive work on the charge and potential energy decreases.

C. The field does negative work on the charge and potential energy increases.

D. The field does negative work on the charge and potential energy decreases.

E. The field does no work on the charge and potential energy increases.

5 μC v E


iClicker: Work and Potential energy

Note: in all those formulas you have to watch your limits of integration / signs. Let’s see about your intuition!The field points left.The 5 μC charge moves right.Which of the following is correct?

A. The field does positive work on the charge and potential energy increases.

B. The field does positive work on the charge and potential energy decreases.

C. The field does negative work on the charge and potential energy increases.

D. The field does negative work on the charge and potential energy decreases.

E. The field does no work on the charge and potential energy increases.

5 μC v E

E points “downhill”, so the potential energy for q>0 increases. Going “uphill” slows the charge down, so the work done by the field is negative.


Force / Work / Potential Energy / Field / Potential for Two Point Charges

Work done by electric force

Electric Force

Set Potential energy = 0 at ∞

Electric fieldJust as the field = the force / q, the potential = the potential energy / q

integrate

differentiate

divide by q

divide by q

• Let’s put a charge Q at the origin, generating an E field, and move a charge q around.

�Efrom Q =�Fon q

q=

kQ

r2r

�Fon q = q �E =kQq

r2r

∆V = −� b

a

�E · d�l = (kQ

rb− kQ

ra)

∆U = −� b

a

�Fc · d�l = (kQq

rb− kQq

ra)

W =� b

a

�Fc · d�l = (kQq

ra− kQq

rb)

U = −� a

∞�Fc ·−d�r =

kQq

ra

�F = −∇U

integrate

differentiate�E = −∇V


More than Two Point Charges• To get the total potential energy of a charge q0 interacting with several other charges, add up the several potential energies

U0 =n�

i=1,...

kq0qi

r0i= kq0

n�

i=1,...

qi

r0i

• To get the total potential energy of the entire system of charges, add up the potential energies of all the pairs of charges

U0 =�

i<j

kqiqj

rij

We need i<j, or we will count each pair twice, or even get ∞’s from ri2/0!


Potential Energy of a 3-charge System

A charge -q is brought from ∞ so that it and two fixed charges, of ±q, form an equilateral triangle.What is the potential energy of the third charge? What is the potential energy of the system?

A. U3 = 2kq2/d, Usys=6kq2/d.

B. U3 = 0, Usys=kq2/d.

C. U3 = 2kq2/d, Usys=3kq2/d.

D. U3 = 0, Usys=-kq2/d.

E. U3 = 0, Usys=-2kq2/d.

-q

-q

q -q



A charge -q is brought from ∞ so that it and two fixed charges, of ±q, form an equilateral triangle.What is the potential energy of the third charge? What is the potential energy of the system?

A. U3 = 2kq2/d, Usys=6kq2/d.

B. U3 = 0, Usys=kq2/d.

C. U3 = 2kq2/d, Usys=3kq2/d.

D. U3 = 0, Usys=-kq2/d.

E. U3 = 0, Usys=-2kq2/d.

-q

-q

q -qOf the 3 pairs, 2 are +- with U<0, and 1 is -- with U>0, so Utotal = -kq2/d. The 3rd charge interacts with + and -, so U3 = 0.



Do these two systems have the same, or opposite, system potential energies?One is +q, -q, -q, the other is +q, +q, -q.

-q

-q

q -q

-q

-q

q q

In both cases, we have 2 +- pairs with U<0, and 1 same sign pair U>0, so Utotal = -kq2/d is the same.


Thank you, andSee you next Thursday


physics 227: lecture 5 gaussâ€™s law leftovers, work

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