Example: 2 flat sheets
- Sheets are parallel, 0.87 cm apart, and have area 120cm2
- Left sheet has charge = -9.3 nC, right has charge = +9.3 nC
- A proton is placed between two sheets, at rest, 1.0 mm from right
- What change in potential does the proton experience as it travels to the left sheet, and what is its speed when it reaches left?
Solution
- Step 1: Context (sheets + charge, speed → charge will experience force that accelerates it, we can use E field to find force and acceleration and then speed, or use potential and energy to find speed)
- Step 3: ... Looking for vf
- Step 4: Potential of infinite sheet, conservation of mechanical energy (kinetic and electric potential energy)
- Step 5:
ΔVleft = -2πkeσleftΔdleft
ΔVright = -2πkeσrightΔdright
σ = Q / A
ΔKE + ΔPEe = 0
PE = qtest V
KE = 1/2 m v2
- Step 6:
1) Find σ for each sheet
2) Find ΔVleft and ΔVright
3) Find ΔPE
4) Find ΔKE
5) Find vf
- Step 7:
σleft = -7.8×10-7 C/m2, σright = 7.8×10-7 C/m2
VLi = -2πke(-7.8×10-7 C/m2)(0.0077 m) = 337 V
VLf = 0
VRi = -2πke(-7.8×10-7 C/m2)(0.0010 m) = -44 V
VRf = -2πke(-7.8×10-7 C/m2)(0.0087 m) = -381 V
ΔVL = 0 V - 337 V = -337 V
ΔVR = -381 V - (-44 V) = -337 V
ΔPE = qtest (ΔVL + ΔVR) = 1.6×10-19 C × -674 V = -1.1×10-16 J
ΔKE = 1.1×10-16 J = 1/2 m vf2
vf2 = 2*1.1×10-16 J / m = 1.3×1011 (m/s)2
vf = 3.6×105 m/s
- Step 8: speed must be in m/s, speed must be less than 3×108 m/s, and speed is a magnitude, so it must be positive.
Notes: Capacitor
- Any device onto which opposite charges can be added / removed so that they will remain separated
- Ex: Two flat sheets of conductor (parallel to each other)
- Counter Ex: One block of conductor (what happens if you charge opposite sides?)
- Net charge on a capacitor will ALWAYS be 0 (Q = charge on a single side, not net)
- Capacitors are used to store energy
Capacitance
- Q = C ΔV (C = capacitance)
- C units = Farads (F) = C / V or C2 / J
- For parallel plate, C = A/(4πd)
- C = 10-12 - 10-6F for small parallel plates, 10-6-10F for common electronics
- C does NOT depend on charge, or energy.
- Greater capacitance means more charge can be separated at a low potential.
Energy
Cautions
- Beware! C the variable = Capacitance. C the unit = Coulombs
- V the variable = electric potential. V the unit = Volts
Problem: Capacitor
- How much energy was stored in the capacitor above (A = 100cm2, d = 1.0cm, Q = 12nC)
Solution:
- C = 8.8×10-12
- PE = 1/2 C (ΔV)2 = 1/2 Q2/C
- PE = 8.1×10-6J