Coulomb’s Law

• • • • • • • • • Coulomb’s Law Electric Field Electric Field from Multiple Charges Integration of Volume charge Electric Field near Infinite Wire Electric Field near Infinite Sheet Electric Field between two Infinite Sheets Field Lines Streamlines

Coulomb’s Law

• Coulomb’s Law with k = 9 x 10 9 Nm 2 /C 2 ε o = 8.85 x 10 -12 C 2 / Nm 2 • 𝐹 = 𝑘 𝑄 1 𝑄 2 𝑅 2 = 𝑄 1 𝑄 2 4𝜋𝜀 𝑜 𝑅 2 Unit vector from r 1 to r 2 𝒂 𝟏𝟐 = 𝒓 𝟐 − 𝒓 𝟏 𝒓 𝟐 − 𝒓 𝟏 • Combining 𝐹 = 𝑄 1 𝑄 2 4𝜋𝜀 𝑜 𝑅 12 2 𝒂 𝟏𝟐 = 𝑄 1 𝑄 2 4𝜋𝜀 𝑜 𝑅 12 2 𝒓 𝟐 − 𝒓 𝟏 𝒓 𝟐 − 𝒓 𝟏 • (Action reaction F 1 = -F 2 )

Example of Coulomb’s Law

• • Force of charge 1 on charge 2 – Charge 1 3 x 10 -4 C at M(1,2,3) – Charge 2 -1 x 10 -4 C at N(2,0,5) • Coulomb’s Law 𝐹 2 = 𝑄 1 𝑄 2 4𝜋𝜀 𝑜 𝑅 12 2 𝒓 𝟐 𝒓 𝟐 − 𝒓 𝟏 − 𝒓 𝟏 R magnitude 𝑅 12 = 𝒓 𝟐 − 𝒓 𝟏 = 2 − 1 𝒂 𝒙 + 0 − 2 𝒂 𝒚 + 5 − 3 𝒂 𝒛 = 3 • • Unit vector 𝒂 𝟏𝟐 = 𝒓 𝟐 − 𝒓 𝟏 𝒓 𝟐 − 𝒓 𝟏 = 𝒂 𝒙 − 2𝒂 𝒚 + 𝟐𝒂 𝒛 𝟑 Result 𝐹 2 3 × 10 −4 𝐶 − 1 × 10 −4 𝐶 = 4𝜋 8.85 × 10 −12 𝐶 2 𝑁𝑚 2 (3 𝑚) 2 𝒂 𝒙 − 2𝒂 𝒚 + 𝟐𝒂 𝒛 𝟑 = − 10𝒂 𝒙 + 20𝒂 𝒚 − 20𝒂 𝒛 𝑵

Electric Field

• Electric Field – – – Coulomb’s Law without 2 nd charge Separates Problem into “Background” and “Test Charge” Units newtons/coulomb (volts/meter) 𝑭 𝒕 𝑬 = 𝑄 𝑡 = 𝑄 4𝜋𝜀 𝑜 𝑅 2 𝒂 𝑹 𝑄 = 4𝜋𝜀 𝑜 𝑅 2 𝒓 − 𝒓′ 𝒓 − 𝒓′ – For source charge at r’ observed at

r’

𝑬 𝒓 = 4𝜋𝜀 𝑜 𝑄 𝒓 − 𝒓′ 2 𝒓 − 𝒓′ 𝒓 − 𝒓′ 𝑥 − 𝑥′ 𝒂 𝒙 = 4𝜋𝜀 𝑜 𝑥 − 𝑥′ + 𝑦 − 𝑦′ 𝒂 𝒚 2 + 𝑦 − 𝑦′ 2 + 𝑧 − 𝑧′ 𝒂 𝒛 + 𝑧 − 𝑧′ 2 3 2 • For source charges at r 1 and r 2 , observed at r 𝑬 𝒓 = 4𝜋𝜀 𝑜 𝑄 1 𝒓 − 𝒓 𝟏 2 𝒓 − 𝒓 𝟏 𝒓 − 𝒓 𝟏 + 4𝜋𝜀 𝑜 𝑄 2 𝒓 − 𝒓 𝟐 2 𝒓 − 𝒓 𝟐 𝒓 − 𝒓 𝟐

Electric Field from Multiple Charges

• 2 source charges at 1 and 2, observed at r 𝑬 𝒓 = 4𝜋𝜀 𝑜 𝑄 1 𝒓 − 𝒓 𝟏 2 𝒂 𝟏 + 4𝜋𝜀 𝑜 𝑄 2 𝒓 − 𝒓 𝟐 2 𝒂 𝟐 • • • Multiple source charges at m, observed at r 𝑛 𝑬 𝒓 = 𝑚=1 4𝜋𝜀 𝑜 𝑄 𝑚 𝒓 − 𝒓 𝒎 2 𝒂 𝒎 Infinite # source charges, observed at r 𝑬 𝒓 = 𝜌 𝑣 4𝜋𝜀 𝑜 𝒓 ′ 𝑑𝑣′ 𝒓 − 𝒓′ 2 𝒂′ We’re going to spend some time on the last one!

Example – Electric Field from 4 charges

• • Sources charges at P 1 (1,1,0), P 2 (-1,1,0), P 3 (-1,-1,0), P 4 (1,-1,0). Each 3 nC.

Observation point r at P(1,1,1) – P 1 (1,1,0) 𝒓 − 𝒓 𝟏 = 𝟏 – P 2 (-1,1,0) 𝒓 − 𝒓 𝟐 = 5 – P 3 (-1,-1,0) 𝒓 − 𝒓 𝟑 = 𝟑 𝒓−𝒓 𝟏 𝒓−𝒓 𝟏 = 𝒂𝒛 1 𝒓−𝒓 𝟐 𝒓−𝒓 𝟐 = 𝟐𝒂 𝒙 +𝒂 𝒛 5 𝒓−𝒓 𝟑 𝒓−𝟑 = 𝟐𝒂 𝒙 +𝟐𝒂 𝒚 +𝒂 𝒛 𝟑 – P 4 (1,-1,0) 𝒓 − 𝒓 𝟒 = 5 𝒓−𝒓 𝟒 𝒓−𝟒 = 𝟐𝒂 𝒚 +𝒂 𝒛 5 • Total field is: 3 × 10 −9 𝐶 𝑬 = 4𝜋 8.85 × 10 −12 𝒂 𝒛 1 ∙ 1 + 𝟐𝒂 𝒙 + 𝒂 5 ∙ 5 𝒛 + 𝟐𝒂 𝒙 + 𝟐𝒂 𝒚 9 ∙ 3 + 𝒂 𝒛 + 𝟐𝒂 𝒚 + 𝒂 5 ∙ 5 𝒛 𝐸 = 6.82𝒂 𝒙 + 6.82𝒂 𝒚 + 32.8𝒂 𝒛

Continuous Charge - Integration of Charge

• • • • • Differential charge element 𝑑𝑄 = 𝜌 𝑉 𝑑𝑉 Integrate for total charge 𝑄 = 𝑉 𝜌 𝑉 𝑑𝑉 Example – – charge density 𝜌 𝑉 = −5 × 10 −6 𝑒 −10 5 𝜌𝑧 𝐶/𝑚 2 Find total charge over region 0 <ρ<1 cm, 2cm

2𝜋 0.01

𝑄 = −5 × 10 −6 𝑒 −10 5 𝜌𝑧 𝜌 𝑑𝜌 𝑑𝜑 𝑑𝑧 0.02

0 0

Integration of Charge (cont)

• Integration on φ 0.04

0.01

𝑄 = −10 −5 𝜋 𝑒 −10 5 𝜌𝑧 𝜌 𝑑𝜌 𝑑𝑧 0.02

0 • Integration on z 0.01

𝑄 = 0 −10 −10 −5 5 𝜌 𝜋 𝑒 −10 5 𝜌𝑧 .02

.04

𝜌 𝑑𝜌 (𝑚𝑢𝑙𝑡𝑖𝑝𝑙𝑦 𝑏𝑦 2𝜋) (𝑑𝑖𝑣𝑖𝑑𝑒 𝑏𝑦 − 10 5 𝜌 𝑐𝑎𝑛𝑐𝑒𝑙) 0.01

𝑄 = 10 −10 𝜋 0 𝑒 −4000𝜌 − 𝑒 −2000𝜌 𝑑𝜌 = 10 −10 𝜋 𝑒 −4000𝜌 − −4000 𝑒 −2000𝜌 −2000 0 .01

𝑄 ≈ −10 −10 𝜋 1 −4000 − 1 −2000 (𝑠𝑒𝑡𝑡𝑖𝑛𝑔 𝑒 −2000∗.01

≈ 0) 𝑄 = −10 −10 𝜋 1 2000 − 1 4000 = −10 −12 𝜋 40 = −0.0785 𝑝𝐶

•

Continuous Charge Other examples

Setup Cartesian – – – – 0.1 ≤ 𝑥 , 𝑦 , 𝑧 ≤ 0.2, 𝜌 𝑉 = 1 𝑥 3 𝑦 3 𝑧 3 Integrate volume −0.2 𝑡𝑜 + .02

Subtract volume −0.1 𝑡𝑜 + .0

1 Q will be zero from integration of odd function.

• Setup cylindrical – 0 ≤ 𝜌 ≤ 0.1, 0 ≤ 𝜑 ≤ 𝜋, 2 ≤ 𝑧 ≤ 4, 𝜌 𝑉 = 𝜌 2 𝑧 2 sin(0.6𝜑 ) – Differential volume 𝜌 𝑑𝜌 𝑑𝜑 𝑑𝑧 • Universe – 𝜌 𝑉 = 𝑒 −2𝑟 /𝑟 2 2𝜋 𝑄 = 0 0 𝜋 0 ∞ 𝑒 −2𝑟 /𝑟 2 𝑟 2 ∞ 𝑠𝑖𝑛𝜃 𝑑𝑟𝑑𝜃𝑑𝜑 = 2𝜋 ∙ 2 𝑒 −2𝑟 𝑑𝑟 = − 4𝜋𝑒 0 −2 0 = 6.28 𝐶

•

Continuous Charge Middle Example

Integral is 4 𝜋 0.1

𝑄 = 𝜌 2 𝑧 2 sin 0.6𝜑 𝜌 𝑑𝜌 𝑑𝜑 𝑑𝑧 2 0 0 𝑄 = 𝜌 4 4 0 .1

𝑧 3 3 2 4 0 𝜋 sin 0.6𝜑 𝑑𝜑 𝑄 = 2.5 × 10 −5 𝜋 56 3 sin 0.6𝜑 𝑑𝜑 0 𝑄 = 2.5 × 10 −5 56 3 − cos(0.6𝜑) 0.6

0 𝜋 = 1.018 𝑚𝐶

Continuous Charge -Field near infinite line charge

• • Will do in cylindrical coordinates – – – Observation on y axis, z = 0 plane 𝒓 = y𝒂 𝒚 Source distributed along z axis 𝒓′ = 𝒛′𝒂 𝒛 Linear charge density constant 𝜌 𝐿 = 𝜌𝒂 𝝆 Source to observation vector 𝐑 = 𝒓 − 𝒓′ = 𝜌𝒂 𝝆 − 𝒛′𝒂 𝒛 • Differential Field Contribution 𝒅𝑬 = 𝜌 𝐿 𝑑𝑧 ′ (𝒓−𝒓 ′ ) 4𝜋𝜀 𝑜 𝒓−𝒓 ′ 3 𝒅𝑬 = 𝜌 𝐿 𝑑𝑧 ′ (𝜌𝒂 𝝆 −𝑧 ′ 𝒂 𝒛 ) 3 4𝜋𝜀 𝑜 𝜌 2 +𝑧 ′2

Field near infinite line of charge (cont)

• ρ and z components 𝑑𝐸 𝜌 = 𝜌 𝐿 4𝜋𝜀 𝑜 𝑑𝑧 ′ (𝜌𝒂 𝝆 ) 𝜌 2 +𝑧 ′2 3 2 𝑑𝐸 𝑧 = −𝜌 𝐿 4𝜋𝜀 𝑜 𝑑𝑧 ′ ( 𝑧 ′ 𝒂 𝒛 ) 𝜌 2 +𝑧 ′2 3 2

(odd - integrates to zero)

• Integration for a long wire is thus 𝐸 𝜌 ∞ = −∞ 4𝜋𝜀 𝑜 𝜌 𝐿 𝜌 2 𝜌𝑑𝑧 ′ + 𝑧 ′2 3 2 𝐸 𝜌 = 𝜌 𝐿 4𝜋𝜀 𝑜 𝜌 1 𝜌 2 𝑧′ 𝜌 2 + 𝑧′ 2 −∞ ∞ = 𝜌 𝐿 4𝜋𝜀 𝑜 𝜌 1 − −1 = 𝜌 𝐿 2𝜋𝜀 𝑜 𝜌 𝑬 = 𝜌 2𝜋𝜀 𝐿 𝑜 𝜌 𝒂 𝝆

•

Field near infinite sheet of charge

Given an infinite line charge and surface density ρ s 𝐸 𝜌 = 𝜌 𝐿 2𝜋𝜀 𝑜 𝜌 • x and y components 𝜌 𝑆 𝑑𝐸 𝑥 = 𝑑𝑦′ 𝑐𝑜𝑠𝜃 2𝜋𝜀 𝑜 𝑥 2 + 𝑦 2 𝑑𝐸 𝑦 = 𝜌 𝑆 2𝜋𝜀 𝑜 𝑑𝑦′ 𝑥 2 +𝑦 2 𝑠𝑖𝑛𝜃

(odd - integrates to zero)

𝑑𝐸 𝑧 = 0

(symmetry)

• x component 𝑑𝐸 𝑥 = 2𝜋𝜀 𝜌 𝑆 𝑑𝑦′ 𝑜 𝑥 2 +𝑦 2 𝑥 𝑥 2 +𝑦 2 = 𝜌 𝑆 𝑥 𝑑𝑦′ 2𝜋𝜀 𝑜 𝑥 2 +𝑦 2

Field near infinite sheet (cont)

• Integration for a sheet is thus ∞ 𝑬 = 2𝜋𝜀 𝑜 𝜌 𝑠 𝑥𝑑𝑦′ 𝑥 2 + 𝑦 ′2 −∞ 𝑬 = 𝒂 𝒙 𝜌 𝑠 2𝜋𝜀 𝑜 𝑡𝑎𝑛 −1 𝑦′ 𝑥 −∞ ∞ 𝒂 𝒙 𝑬 = 𝜌 𝑠 2𝜋𝜀 𝑜 𝜋 2 − −𝜋 2 𝒂 𝒙 𝑬 = 𝜌 𝑠 2𝜀 𝑜 𝒂 𝒙 • Field points away +𝒂 𝒏 +𝑄 , toward −𝒂 𝒏 −𝑄 • Field is independent of distance r<

Electric Field between 2 Infinite Sheets

(-Q) charge sign and unit vector reversed) 𝑬 = 𝜌 𝑠 2𝜀 𝑜 𝒂 𝒙 + −𝜌 𝑠 2𝜀 𝑜 −𝒂 𝒙 = 𝜀 𝜌 𝑠 𝒂 𝒙 𝑜

Field Lines

•

Field lines

– Point in direction of electric field – Direction + test charge moves – Originates on +Q terminates on -Q – Cross-sectional density proportional to E magnitude

•

Streamlines

Equation of line which follows field line at x, y, z – – slope of this line y=f(x) 𝑑𝑦 𝑑𝑥 should equal field ratio 𝐸 𝑦 𝐸 𝑥 – Set 𝑑𝑦 𝑑𝑥 = 𝐸 𝑦 𝐸 𝑥 – Solve for equation y=f(x) as function of x

Streamlines

• • Vector field are A x , A y , and A z From geometry 𝑑𝑦 = 𝐸 𝑦 𝑑𝑥 𝐸 𝑥 function of x,y,z • Example 𝐸 = 𝑥 𝑥 2 +𝑦 2 𝒂 𝒙 + 𝑦 𝑥 2 +𝑦 2 𝒂 𝒚 • Plugging in 𝑑𝑦 𝑑𝑥 = 𝐸 𝑦 𝐸 𝑥 = 𝑦 𝑥 • • Result 𝑙𝑛𝑦 = 𝑙𝑛𝑥 + 𝑙𝑛𝐶 𝑦 = 𝐶𝑥 Plug in x and y at particular point to evaluate C 𝑑𝑦 𝑦 = 𝑑𝑥 𝑥

Streamline Example

• • • • Find streamlines of following in rectangular coordinates 𝑬 = 1 𝜌 𝒂 𝒓 Transforming to rectangular 1 𝑬 = 𝑥 2 + 𝑦 2 𝒂 𝒓 ∙ 𝒂 𝒙 𝒂 𝒙 + 1 𝑥 2 + 𝑦 2 𝒂 𝒓 ∙ 𝒂 𝒚 𝒂 𝒚 𝑬 = 𝑥 2 1 + 𝑦 2 𝑐𝑜𝑠 𝜑 𝒂 𝒙 + 𝑥 2 1 + 𝑦 2 𝑠𝑖𝑛(𝜑) 𝒂 𝒚 𝑬 = 𝑥 2 𝑥 + 𝑦 2 𝒂 𝒙 + 𝑥 2 𝑦 + 𝑦 2 𝒂 𝒚 Plugging in streamline equation 𝑑𝑦 = 𝑑𝑥 𝐸 𝑦 = 𝐸 𝑥 𝑦 𝑥 𝑑𝑦 = 𝑦 𝑑𝑥 𝑥 Solution 𝑙𝑛𝑦 = 𝑙𝑛𝑥 + 𝑙𝑛𝐶 𝑦 = 𝐶𝑥 • At P(-2,7,10) y = -3.5 x

Example problem 1

1. 3 point charges are in xy plane; with 5 nC at y= 5 cm, -10 nC at y =-5 cm, and 15 nC at x=-5cm. Find position of 20 nC that exactly cancels field at origin.

- Add first 3 fields to get resultant as function of a

x

, a

y

(like example 2.2) - 4 th charge must exactly cancel field with same combination of a

x

, a

y

- Write in general field form as magnitude times unit vector - Equate magnitudes

Example problem 2

7.

A 2uC charge is located at A(4,3,5) in free space. Find E ρ , E φ , and E z P(8,12,2) at - Get field in rectangular coordinates as function of a

x

, a

y , a z

- translate rectangular variables to cylindrical variables - translate rectangular unit vectors to cylindrical variables.

Example problem 3

13. A uniform charge density extends throughout a spherical shell from r=3 cm to r=5 cm. Find the total charge and the radius containing half the charge.

Example problem 4

• Find the electric field on the z-axis produced byan annular ring z= 0, a <ρ