Physics 227: Lecture 3 Electric Field Calculations ... · PDF filePhysics 227: Lecture 3 Electric Field Calculations, Symmetry, Charged Particle Motion, Flux, Gauss’s Law ... ``Cheat’’

Physics 227: Lecture 3Electric Field Calculations, Symmetry,

Charged Particle Motion, Flux, Gauss’s Law• Lecture 2 review:

• F1 on 2 = k q1 q2 / r2

• Superposition: Forces are vectors, Coulomb forces add as vectors.

• Electric field determined by test charge: E = Fon qtest / qtest

• Electric fields are vectors, and add as vectors. Unique direction and magnitude at each point in space.

• Efrom point charge = k q / r2

• Field lines show direction of electric field: E tangent to find line, fields lines start on + charge, end on - charge, but may start / stop at r = ∞, field strength ∝ number of field lines, number of field lines ∝ charge.

``Cheat’’ / formula sheet added to Sakai resources - tell me about any typos

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Monday, September 12, 2011

• Superposition allows the electric field to be calculated for an arbitrary fixed charge distribution ρ(x,y,z). Let:

Calculating the Electric Field

�E(�x�) =�

i

qi

4π�0r2i

ri →�

d3xρ(�x)

4π�0|�x− �x�|2 ∆

• Or:

�E(�x�) =�

i

qi

4π�0r2i

ri →�

dq1

4π�0r2∆

• Where Δ is the unit vector from the position of the charge to the position at which we are evaluating the electric field.

• The integrals might require numerical evaluation, turning this into a computation problem.


Electric Field of Two Charges

�E(�x�) =�

i

qi

4π�0r2i

ri →�

d3xρ(�x)

4π�0|�x− �x�|2 ∆

• What is the electric field along the x axis?

• For x > d:

• For 0 < x < d:

q1q2x

y

dd

�E =q1

4π�0|x− d|2 x +q2

4π�0|x + d|2 x

�E =−q1

4π�0|x− d|2 x +q2

4π�0|x + d|2 x

Notice that at x >> d the

field falls like the field of a point charge



�E(�x�) =�

i

qi

4π�0r2i

ri →�

d3xρ(�x)

4π�0|�x− �x�|2 ∆

• What is the electric field along the y axis?

• You can see the vertical components will cancel.

q1-q1x

y

dd

E

�E =−2q1

4π�0(y2 + d2)x× d�

y2 + d2

�E =−2q1d

4π�0(y2 + d2)3/2x

Notice that at large y the dipole field falls faster than the field of a point charge



q1-q1x

y

dd

E

• Why does the field fall faster than the field of a point charge?

A. It cannot - the professor is wrong!

B. It always falls faster for 2 or more charges.

C. The two charges cancel, so the field is 0 everywhere!

D. Because the fields of the two charges partially cancel.

E. It falls faster in some directions, but not all.

�E =−2q1d

4π�0(y2 + d2)3/2x




q1-q1x

y

dd

E

�E =−2q1d

4π�0(y2 + d2)3/2x


• Why does the field fall faster than the field of a point charge?

A. It cannot - the professor is wrong!

B. It always falls faster for 2 or more charges.

C. The two charges cancel, so the field is 0 everywhere!

D. Because the fields of the two charges partially cancel.

E. It falls faster in some directions, but not all.

A. Often, but not this time.B. Not for, e.g., two same

sign charges.C. There is a partial, not

total, cancellation.E. Dipole field falls like 1/r3, though you have not been told enough so far to know

this answer is wrong Monday, September 12, 2011

Electric Field of a Ring of Charge• What is the electric field

along the x axis?

• You can see the vertical components will cancel.

• The integrand does not depend on position along the ring, so the integral can be done ``by inspection’’

Notice that at large x >> a, this approaches the field of a point charge.

�E(�x�) =�

i

qi

4π�0r2i

ri →�

dq1

4π�0r2∆

Example 21.10


Symmetry Arguments

The charge looks the same to both observers. The charge looks the same when an observer rotates about its head. There is nothing to distinguish the observer’s up from your down, or left from right. There is no preferred direction in space except for the line going from the charge to the observer. The field must be

along this line. (You are a point in space, your body does not count!)

Sometimes you can determine something about the shape of the electric field from symmetry arguments

For a point charge or a uniform sphere of charge, the field is radial and spherically symmetric: E(r,θ,φ) → E(r)

+ +

+


Symmetry Arguments

If you move in z, the line looks the same, so the field has no z component. If you go around the line in the φ direction, the line looks the same, so the field

has no φ component. The field can only be in the ρ direction.


For an infinitely long uniform line of charge, the field is radial and perpendicular to the line: E(ρ,φ,z) → E(ρ)

+

+

+ + + + + + + + + + +

ρ

z

φ


Symmetry Arguments

If you move in x or y, or if you rotate about the z axis, the plane looks the same, so the field has no x or y components. The field can only be in the z direction.


For an infinite uniform plane of charge, the field is perpendicular to the plane: E(x,y,z) → E(z)

+

+ +

+ +

+ +

+ +

+

+ +

+ +

+ +x

y

+++


Motion in an Electric Field• Soon we will learn that the field inside a ``parallel plate capacitor’’ is ≈ constant, as long as the size of the plates (√A) is much greater than the separation of the plates (d)

• Since the E-field is constant, the electric force F = qE is constant between the plates.

+ v0

• Thus, F = ma leads to ay = qE/m and ax = 0.• With v(t=0) = v0x, we find that x = v0xt, and y = (1/2)at2 = (1/2)(qE/m)t2.• Put t = x/v0x into the equation for y to obtain: y = (1/2)(qE/mv0x2)x2.• This is a parabolic trajectory, just like motion of a projectile in a uniform gravitational field. It does not follow the field lines.

y

x


Introduction to Flux

• For a constant E-field, flux through a surface is, crudely speaking, the number of field lines through the surface.

• Toy problem: inside the parallel plate capacitor the E field is constant and vertical, 25 field lines per m2

• There is a 1 m2 horizontal surface in side the capacitor.• What is the flux through this horizontal surface?• Flux = field line density x area = 25 field lines/m2 x 1 m2 = 25 field lines


Toy Examples, with flux of water

• Water of density 1000 kg/m3 flows at 3 m/s through a horizontal pipe of cross sectional area 1 m2. What is the flux of water through a vertical plane through the pipe?• Flux = 1000 kg /m3 x 3 m/s x 1 m2 = 3000 kg/s.

• Water of density 1000 kg/m3 flows at 3 m/s through a horizontal pipe of cross sectional area 1 m2. What is the flux of water through a horizontal plane through the pipe?• Flux = 0. No water goes through the plane! Water flows along the plane, but it does not cross over it.

Velocity of water is a vector field,

like electric fields.


More toy examples

• Water of density 1000 kg/m3 flows at 3 m/s through a U-shaped horizontal pipe of cross sectional area 1 m2. What is the flux of water through a vertical plane through the upper+lower pipes?• Fluxupper = 1000 kg /m3 x 3 m/s x 1 m2 = 3000 kg/s.• Fluxlower = 1000 kg /m3 x -3 m/s x 1 m2 = -3000 kg/s.• Fluxtotal = Fluxupper + Fluxlower = 0


More formal definition of Flux

ΦE =�

�E · d �A =�

E cos φdA =�

E⊥dA

A = x �E = E0(cos φx + sinφy)

E-field lines

Surface: A unit vector

⊥ to surface

E-field lines

ˆtry

More generally we have a curved surface with�E = Ex(x, y, z)x + Ey(x, y, z)y + Ez(x, y, z)zand the integral will be hard to evaluate.

ΦE =�

E0(cos φx + sinφy) · dA x = E0 cos φA


Closed Surfaces + Gauss’s Law

• We are going to be concerned with the total electric flux through a closed surface.• In advanced integral calculus E&M, you learn that the surface integral of the flux can be related to a volume integral that depends on the charge in the volume.• The book does not show this; we will just accept the result:

ΦE =�

�E · d �A =qenclosed

�0The result does not depend on the shape of the closed surface!


Closed Surfaces

A. The radius of the sphere decreases to a/2.

B. The magnitude of each charge is doubled.

C. The sphere moves sideways so its surface goes through the origin.

D. The sphere deforms into a cube with each side 2a long centered at the origin.

E. The charge doubles in magnitude and the sphere doubles in surface area to compensate.

• There is a charge distribution near the origin.• The charge furthest from the origin is a distance a away.• There is a Gaussian sphere of radius b centered at the origin, with b > a.•In which of the following cases does the flux through the surface definitely not change?•There may be more than one right answer.


Closed Surfaces

A. The radius of the sphere decreases to a/2.

B. The magnitude of each charge is doubled.

C. The sphere moves sideways so its surface goes through the origin.

D. The sphere deforms into a cube with each side 2a long centered at the origin.

E. The charge doubles in magnitude and the sphere doubles in surface area to compensate.

• There is a charge distribution near the origin.• The charge furthest from the origin is a distance a away.• There is a Gaussian sphere of radius b centered at the origin, with b > a.•In which of the following cases does the flux through the surface definitely not change?•There may be more than one right answer.

E. charge definitely changes

A.,B.,C.: charge would usually change, but might

not in special cases


Flux through Closed Surfaces


Closed Surfaces

• Through which surface is the flux greatest?•There may be more than one right answer

A. S1

B. S2

C. S3

D. S4

E. S5

• Six charges are in a plane.• For simplicity here, treat all charges as +/-1 μC, ignore the magnitudes shown.• The intersections of 5 closed surfaces with the plane are shown.


Closed Surfaces

• Through which surface is the flux greatest?• There may be more than one right answer.• q in S1 = 0, q in S2 = +1, q in S3 = +2, q in S4 = 0, q in S5 = +2 (all in μC).

A. S1

B. S2

C. S3

D. S4

E. S5

• Six charges are in a plane.• For simplicity here, treat all charges as +/-1 μC, ignore the magnitudes shown.• The intersections of 5 closed surfaces with the plane are shown.


See you Thursday


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Physics 227: Lecture 3 Electric Field Calculations ... · PDF filePhysics 227: Lecture 3 Electric Field Calculations, Symmetry, Charged Particle Motion, Flux, Gauss’s Law ... ``Cheat’’