Demo Equipment

Notes: Infinite plane

Example: constant acceleration

Step 1
  • Context: charge + sheet + uniform, fast / moving / velocity (kinematics)
Step 3
  • me = 9.1×10-31 kg, e = -1.6×10-19C
  • σ = -3.0 nC / (1.0 cm2)
Step 4
  • Electric field of infinite plane of charge
  • constant acceleration (ONLY for constant E, flat sheets)
  • Newton's 2nd
Step 5
  • E = 2 π ke σ
  • Fx = q Ex
  • Newton;s 2nd → ax = Fx / m
  • constant acceleration relationship between v, a, and x → vx2 - v0x2 = 2 ax Δx
Step 6
  • Find E, then F, then a
  • Use a and Δx to find x.
Step 7
  • E = 1.7×106N/C
  • F = 2.7×10-13N
  • ax = 3.0×1017m/s2
  • vx = 2.4×107m/s
Step 8
  • Looking for speed, so units should be m/s
  • Expected range for speeds? Speeds must be lower than 3×108 m/s, and otherwise are dependent on context (car can not move 107 m/s, but electron can.
  • Sign: Electron moves away from sheet, so if sheet is to left of the electron, vx must be positive.

Example: parallel sheets

Field due to left sheet
  • E = 5650N/C
  • to the left of sheet, E points right...
Field due to right sheet
  • E = 5650N/C
  • to the left of sheet, E points left...
Total field
  • to left, E = 0
  • between, E = 11300N/C = 4πkeσ
  • to right, E = 0
Depiction

Demo: conductor in a uniform electric field

In uniform field (to right)
  • Conductor becomes polarized.
  • -q on left side
  • +q on right side
  • F on left charge = qE to left
  • F on right charge = qE to right
  • Net force = 0
Wimshurst
  • turning crank causes charge separation (one jar becomes + charged, other becomes - charged)
  • turn crank 1" spark gap → 1" sparks
  • sparks → E field is high enough to pull electrons off of air molecules.
Demo 1
  • Connect opposite jars to conducting sheets
  • turn crank → opposite charged sheets
  • Conducting can hanging between experiences no net force
Demo 2
  • If I take charge off of the left sheet and put it on the can, what happens?
  • Charge diagram (include charge separation, plus excess charge)
  • Turn crank, for fun

Notes: Uniform sphere

Example
  • Imagine there is a hollow spherical conductor: r=15cm. Q = +450nC
  • Q: If I put a +1 nC charge 0.50cm from the left inner wall, what force does the charge feel?
  • A: no force!
  • For every bit of charge above pushing down, there is an equal bit of charge below pushing up (easy).
  • There is a small amount of charge nearby to the left, and a LOT MORE charge further away to the right...

Example / Demo

Example
  • Confetti with mass of 0.052 g.
  • Sphere has radius of 6.8 cm, charge of 750 nC.
  • 0.14 nC jumps to the confetti.
  • What is the confetti's acceleration?
  • Step 1 - "on" (normal), charged + sphere (electric field), mass (gravitational force), acceleration (newton's 2nd)
  • Force diagram - with confetti on top, a is not 0 only if FN = 0.
  • Step 3 - Q = 750 nC, q_test = 0.14 nC, R = 0.068 m, m = 0.000052 kg
  • Step 4 - (See 1)
  • Step 5
    E = ke Q / r2
    Fe = q_test E
    Fg = m g
    Fey + Fgy = m ay
  • Step 6 - Solve first for E, then send for Fe, then third for Fg, fourth for ay
  • Step 7
    E = 1.5×106 N/C ← This makes sense, because it is high enough charge is leaking from the sphere.
    Fe = 2.0×10-4 N
    Fg = 5.1&time;10-4 N
    ay = -5.9 m/s2
  • Step 8 - units should be m/s2
    No expected magnitude range for accelerations
    Because the y component is negative, the confetti would accelerate downwards. This is not possible because there is a sphere in the way. This only shows that the electric force is not strong enough to lift the confetti, and the acceleration is actually 0.
    • If the above calculation is repeated with q = 3.9×10-10 C, the acceleration would be 1.1 m/s2