Electrical Energy and Capacitors


Outline:

  • Electric Potential
  • Va = Ua/q
  • [[#|Potential]] Difference/Voltage
  • measured in volts
  • Vba = - Wba/q
  • Electric Field
  • Vba = - Ed
  • Point Charges
  • E = (1/4πε0) (Q/r^2)
  • V=(1/4πε0) (Q/r)
  • Equipotential Lines
  • all at the same potential, perpendicular to the electric field
  • Electrostatic Potential Energy
  • U = qVba
  • U = (1/4πε0) (Q1Q2/r)
  • The Rutherford Experiment
  • Capacitance
  • Q = CVba
  • E = Q/ε0A
  • Of a Parallel Plate Capacitor
  • C = (ε0) (A/d)
  • Of a Cylindrical Capacitor
  • C = 2πε0L/ln(Ra/Rb)
  • Of a Spherical Capacitor
  • C = 4πε0(rarb/ra - rb)
  • In Parallel Circuit
  • Q = Q1 + Q2 + Q3
  • C = C1 + C2 + C3
  • In Series Circuit
  • V = V1 + V2 + V3
  • 1/C = 1/C1 + 1/C2 + 1/C3
  • Dielectrics
  • non-conducting space between conductors
  • K = dielectric constant
  • C = (Kε0)(A/d)
  • Electric [[#|Energy Storage]]
  • U = (1/2)(CV^2)

Investigation:


We mapped out the electric field and equipotential lines using a battery and a multimeter on conductive paper. This helped to show us how equipotential lines interact with the electric field. This is important in order to understand where the electric field changes and flux in general.

Multiple Choice:


1. Derive the expression for the capacitance of a spherical conductor using Gauss's [[#|law]].
a. C = (ε0) (A/d)
b. C = 2πε0L/ln(Ra/Rb)
c. C = 4πε0(rarb/ra - rb)
d. C = -2πε0(rarb/ra - rb)^-3
e. C = 2πε0L/ra - rb

2. Potential difference is know as:

a. Work per unit charge
b. Voltage per unit charge
c. Electric field per unit charge

d. Potential per unit charge
e. Charge per unit charge

3.
external image 00330.png
What is Q equal to for this circuit?
a. 1500 µC
b. 150 µC
c. 1. 5 x 10^ -6 µC
d. 15 µC
e. 1.5 µC

4. What are the charges across each plate for a series circuit where C1 = 2 µF C2 = 4 µF and Vab = [[#|30V]]
a. Q1 = 40 µC Q2 = 40 µC
b. Q1 = 40 µC Q2 = 80 µC
c. Q1 = 80 µC Q2 = 40 µC
d. Q1 = 80 µC Q2 = 80 µC
e. Q1 = 400 µC Q2 = 400 µC

5. for a parallel circuit?
a. Q1 = 60 µC Q2 = 120 µC
b. Q1 = 60 µC Q2 = 60 µC
c. Q1 = 120 µC Q2 = 60 µC
d. Q1 = 120 µC Q2 = 120 µC
e. Q1 = 600 µC Q2 = 1200 µC

Open Response:


1. A 1000 N/C electric field with a distance of 60 cm between points X and Y in a line 10 degrees from the electric field.
a. What is the voltage between X and Y?
b. How much work is done on a proton on the line?
c. What is the final velocity on a proton on the line starting from rest?
d. What is the force on the proton?
e. What is the electric potential energy for a proton on the line?


Horizontal:

Time (s)
Horizontal Position (cm)
1
10
2
25
3
35
4
50
5
70
6
90


AP_Physics_4.png

Analysis of Horizontal Position vs. Time

The above horizontal position vs. time graph [[#|shows]] a curve with a constant slope. Because the slope represents velocity, the velocity is constant.


Calculations of Horizontal Velocity

v = change in position/change in time
From 1s to 2s: (25 cm - 10 cm)/(2 s - 1 s) = 15 cm/s



Time (s)
Velocity (cm/s)
0.5
10
1.5
15
2.5
10
3.5
15
4.5
20
5.5
20

Analysis of Horizontal Velocity vs. Time


The horizontal velocity vs. time table depicts a constant velocity, though slight errors in collecting [[#|data]] cause for the velocity to appear to change. Because the velocity is constant there is no acceleration.

Vertical:


Time (s)
Vertical Position (cm)
1
85
2
80
3
75
4
60
5
40
6
10
AP_Physics_5.png

Analysis of Vertical Position vs. Time

The above vertical position vs. time graph [[#|shows]] a curve with a decreasing slope. Because the slope represents velocity, this decreasing slope indicates that the velocity is changing.


Calculations of vertical Velocity

v = change in position/change in time
From 1s to 2s: (80 cm - 85 cm)/(2 s - 1 s) = 5 cm/s

Time (s)
Velocity (cm/s)
0.5
85
1.5
5
2.5
5
3.5
15
4.5
20
5.5
30


AP_Physics_6.png

Analysis of Vertical Velocity vs. Time

The vertical velocity vs. time graph depicts a changing velocity, though slight errors in graphing cause for the velocity to appear to decrease in the very beginning when it is always increasing. Acceleration is (10 cm/s - 0 cm/s)/(3 s - 2 s) = 10 cm/s/s.


Conclusion

By analyzing the straight horizontal position vs. time graph and the constant velocity vs. time table we can conclude that the marble does not accelerate horizontally. By analyzing the curved vertical position vs. time graph and the increasing velocity vs. time table we can conclude that the marble accelerates vertically. In this case the acceleration was 10 cm/s/s in the opposite [[#|direction]] of velocity. It can therefore be hypothesized that a marble thrown in a projectile with have a constant velocity horizontally with a constant acceleration vertically.