Demo Equipment
- Dielectric breakdown
- Wimshurst - Would I be able to shade the spark, to make it visible?
- Electroscope - to relate to the Wimshurst
- Levitating Pie tins - tins, 1 wire, alligator clamp
Electric Field (E)
- Electric field has a value (vector) at every point in space
- Any charge will generate electric field.
- Any charge located where the electric field is not 0 will experience force.
- Fx = qtest Ex, ...
Properties
- E = ke q / r2 ← causing charge
- E points straight away from a + charge, straight towards a -
- units = N/C
- Typical magnitudes range from 0 to 106 N/C
- Upper limit is due to 'dielectric breakdown' - Air becomes ionized (electrons pulled away from air molecules) and becomes conducting.
Demo dielectric breakdown with electroscope and Wimshurst
- 1) Charge the electroscope by induction
- There are three steps - I bring the conductor close to a second charge to separate the charges on the conductor, I touch to conductor to allow it to become charged, and I touch the conductor to the electroscope to store the conductor's charge on the electroscope.
- 2) Wimshurst does the same 3 step, just really fast
- The conductors on one wheel have some charge, and so act as the second charge to the other wheel. The diagonal bars have brushes which touch the conductors on the other wheel, allowing its conductors to become charged. The sides have brushes which allow the conductors on the other wheel to deposit their charge at the sides.
- Spark is audible, and in the dark, it would be visible. Up close it is visible as a thin thread of light.
Computing (point charge)
- E = ke q / r2 ← causing charge
- Draw schematic with many arrows.
- 1 cm from 1 nC, E = 9.0×104 N/C
- 2 cm from 1 nC, E = 2.3×104 N/C
- 3 cm from 1 nC, E = 1.0×104 N/C
Relation to F
- If you know a charge and what electrical force it experiences, you know the electric field at its location (E=F/q)
- If you know a charge, and what electric field it experiences, you know what electric force is on it (F=qE)
2 charges
- +1.2 nC, 6.4 cm to left of origin.
- +1.2 nC, 6.4 cm to right of origin.
- What is the net electric field 7.5 cm above origin?
- Good steps: 1) Find r of first charge, 2) Find θ of E field due to first charge, 3) Find magnitude, 4) Find components, 5) Repeat 1-4 for second charge, 6) Add components
Left Charge (1)
- 1) r1 = sqrt(0.642 + 0.752) cm = 0.99 cm
- 2) θ1 = atan(0.75/0.64) = 50°
- 3) E1 = ke q / r2 = 9.0×109 N m2/C2 × 1.2 nC / (0.0099 m)2 = 1.1×105 N/C
- E1x = 7.2×104 N/C
- E1y = 8.5×104 N/C
Right Charge (2)
- Do first 2 steps on your own.
- 1) r2 = 0.99 cm
- 2) θ2 = 130°
- Next two steps
- E2 = 1.1×105 N/C
- E1x = -7.2×104 N/C
- E1y = 8.5×104 N/C
net
- Ex = 0!
- Ey = 1.7×105 N/C
- E field points up (90°)
Point out
- ONLY FOR THIS DISTRIBUTION...
- E field on the positive y axis is always upwards (θ = 90°)
- E field on the negative y axis is always downwards (θ = 270°)
- E field at the origin? (0)
Flat plane of charge
- Imagine I add a 1.2 nC charge 1.28 cm to the left of origin, and another 1.28 cm to the right of the origin
- x components of net field will still cancel along the y-axis
- y components of E fields due to new charges will be 1/4 previous, and add on
- Keep adding evenly spaced charges along the x-axis, and the y component of E will grow increasingly slowly
Uniformly charged plane
- E points straight away from a positively charged plane, straight towards a negatively charged plane
- E = 2πke|σ|
- σ = Q / A (charge per unit area, C/m2)
- Note - E does not depend on distance from the plane!
- Depict idealization (use both many arrows and field lines)
- Depict realistic version. Approximation is best close to plane, relatively far from edge.
Demo - levitating pie tins
- Two identical pie tins. Bottom one is connected to one side of the Wimshurst.
- When the Wimshurst is cranked, eventually the top one levitates off.
- Assume mass of each is 5 g, radius = 13 cm
- How much charge was on the pie tins? Approximate the tins as being flat.
- Step 1: Context (charge + radius→Use electric field, mass)
- Step 2: Schematic, labels (m, r)
- Step 3: Looking for q
- Step 4: Electric field of an infinite plane, Newton's 2nd, vectors (components, addition), gravitational force
- Step 5: Fe = q E ← Electric field
Fg = m g ← gravitational force
E = 2πkeσ, σ = Q / A ← Electric field of infinite plane
Fey + Fgy = 0 ← Newton's 2nd
- Step 6: Determine A, Fg
Use Newton's 2nd to find Fe
Use Electric field / force relationship to find E
Use Electric field of infinite plane to find σ
Use σ definition to find Q.
- Step 7:
A = π (0.13 m)2 = 0.053 m2
Fg = 0.005 kg × 9.8 m/s2 = 0.049 N
Fe = 0.049 N = |Q| E → E = 0.049 N / Q
0.049 N / Q = 2 π ke Q / A → Q2 = 0.049 N × 0.053 m2 / (6.28 × 9.0×109 N m2 / C2
→ Q2 = 4.6×10-14 C2
→ Q = 2.1×10-7 C
- Step 8:
Answer is a charge, so we are looking for units of C
Expected range is 10-10 to 10-5 C. Our answer is in that range
Sign: Since the upper tin will have the same charge as the lower, they will repel whether they are both positive, or both negative. We can not tell which sign the charge has.