Wednesday, May 28, 2008
Tuesday, May 27, 2008
Tuesday, May 27
Today we corrected the Moving Charges Worksheet. I'm feeling so generous so here's the answers:
4) a) Fe and Fg
b) Positive charge
c) Fe = Fg
mg = Eq
q = mg/E = (3.2 x 10-15)(9.8) / (19600)
= 1.6 x 10-18 C
5) B = F/IL
= (0.40) / (20.0)(0.15)
= 0.13 T
6) B = F/IL
= (6.0 x 10-5) / (1.5)(0.50)
= 8.0 x 10-5 T
7) F = BIL
= (5.5 x 10-5)(20.0)(10.0)
= 1.1 x 10-2 N
8) B = F/qv
= (5.6 x 10-13) / (1.6 x 10-19)(2.1 x 105)
= 16.7 T [left]
9) F = Bqv
= (1.2)(1.6 x 10-19)(8.6 x 104)
= 1.65 x 10-14 N [right]
10) B = F/qv
= (4.0 x 10-6) / (1.6 x 10-19)(5.0 x 105)
= 5.0 x 107 T [into the page]
11) q = F/Bv
= (4.8 x 10-14) / (5.0 x 10-2)(2.0 x 106)
= 4.8 x 10-19 C
12)
a) F = Bqv
= (0.080)(1.6 x 10-19)(4.0 x 106)
= 5.12 x 10-14 N
b) R = mv2/F
= (9.11 x 10-31)(4.0 x 106)2 / (5.12 x 10-14)
= 2.84 x 10-4 m
13) v = BqR/m
= (5.0 x 10-4)(1.6 x 10-19)(0.50) / (9.11 x 10-31)
= 4.4 x 107 m/s
We also picked up a Review sheet and a worksheet named "The Path of a Charged Particle in a Magnetic Field." Test on Thursday !
4) a) Fe and Fg
b) Positive charge
c) Fe = Fg
mg = Eq
q = mg/E = (3.2 x 10-15)(9.8) / (19600)
= 1.6 x 10-18 C
5) B = F/IL
= (0.40) / (20.0)(0.15)
= 0.13 T
6) B = F/IL
= (6.0 x 10-5) / (1.5)(0.50)
= 8.0 x 10-5 T
7) F = BIL
= (5.5 x 10-5)(20.0)(10.0)
= 1.1 x 10-2 N
8) B = F/qv
= (5.6 x 10-13) / (1.6 x 10-19)(2.1 x 105)
= 16.7 T [left]
9) F = Bqv
= (1.2)(1.6 x 10-19)(8.6 x 104)
= 1.65 x 10-14 N [right]
10) B = F/qv
= (4.0 x 10-6) / (1.6 x 10-19)(5.0 x 105)
= 5.0 x 107 T [into the page]
11) q = F/Bv
= (4.8 x 10-14) / (5.0 x 10-2)(2.0 x 106)
= 4.8 x 10-19 C
12)
a) F = Bqv
= (0.080)(1.6 x 10-19)(4.0 x 106)
= 5.12 x 10-14 N
b) R = mv2/F
= (9.11 x 10-31)(4.0 x 106)2 / (5.12 x 10-14)
= 2.84 x 10-4 m
13) v = BqR/m
= (5.0 x 10-4)(1.6 x 10-19)(0.50) / (9.11 x 10-31)
= 4.4 x 107 m/s
We also picked up a Review sheet and a worksheet named "The Path of a Charged Particle in a Magnetic Field." Test on Thursday !
Wednesday, May 21, 2008
Today we learned that in order for an object to move from one distance to another distance above the earth's surface, it needs a certain amount of force which increases its gravitational potential energy (work) to raise a mass from d1 to d2.
Electric potential energy is the same thing as when you place a positive charge between parallel plates. The charge is attracted to the negative plate but in order for the plate to move further away a force must be applied upward. Making a second height other than the first. The area of the shaded region between the two heights reps the increase in electric potential energy.
Calculating Electric potential energy.
PE(1)/(force*height1)=(Force)(distance of charge from negative plate)
=?J (amount of electric potential energy the charge has from the -ve plate)
Now you calculate the PE of charge at a second height
PE(2)=(Force)(second distance of charge from negative plate)
=?J(amount of electric potential energy the charge has from the -ve plate at a second distance)
Now to find the total PE b/w the two positions YOU SUBTRACT PE(1) from PE(2). Now the answer doesnt always have to be positive. Only if point B is further away from the negative plate. If point B is closer than it has a loss of PE.
This can be found from the shaded region of the two distances by multiplying the force of both(only if the same force is applied) by the two distances.
So PE is the work that is done in order to move a charge from one place to another.
ELECTRIC POTENTIAL
Electric potential and PE are very different. Electric potential is the potential energy that a charge has at a given point. Which can be found by this equation ( energy,J)/(charge,C)
The quantity is called a volt.
Potential Difference
To calculate the potential difference of a charge once it has moved to another point you must calculate the Electric potential of the first point V(1)=( energy,J)/(charge,C) and the same for the second point V(2)=( energy,J)/(charge,C). Doing the same thing as in Electric potential energy the Voltage of the first point from the voltage of the second point.
DeltaV=V(2)-V(1)
Now the amount of voltage that has changed in the positions of the charge is EQUAL to the amount of work needed to move the charge from one position to the next.
Now we were given a fun worksheet in which we went over in class and a second problem solving sheet which to be finished by tomorrow.
Electric potential energy is the same thing as when you place a positive charge between parallel plates. The charge is attracted to the negative plate but in order for the plate to move further away a force must be applied upward. Making a second height other than the first. The area of the shaded region between the two heights reps the increase in electric potential energy.
Calculating Electric potential energy.
PE(1)/(force*height1)=(Force)(distance of charge from negative plate)
=?J (amount of electric potential energy the charge has from the -ve plate)
Now you calculate the PE of charge at a second height
PE(2)=(Force)(second distance of charge from negative plate)
=?J(amount of electric potential energy the charge has from the -ve plate at a second distance)
Now to find the total PE b/w the two positions YOU SUBTRACT PE(1) from PE(2). Now the answer doesnt always have to be positive. Only if point B is further away from the negative plate. If point B is closer than it has a loss of PE.
This can be found from the shaded region of the two distances by multiplying the force of both(only if the same force is applied) by the two distances.
So PE is the work that is done in order to move a charge from one place to another.
ELECTRIC POTENTIAL
Electric potential and PE are very different. Electric potential is the potential energy that a charge has at a given point. Which can be found by this equation ( energy,J)/(charge,C)
The quantity is called a volt.
Potential Difference
To calculate the potential difference of a charge once it has moved to another point you must calculate the Electric potential of the first point V(1)=( energy,J)/(charge,C) and the same for the second point V(2)=( energy,J)/(charge,C). Doing the same thing as in Electric potential energy the Voltage of the first point from the voltage of the second point.
DeltaV=V(2)-V(1)
Now the amount of voltage that has changed in the positions of the charge is EQUAL to the amount of work needed to move the charge from one position to the next.
Now we were given a fun worksheet in which we went over in class and a second problem solving sheet which to be finished by tomorrow.
Friday, May 16, 2008
Thursday, May 15, 2008
Thursday, May 15, 2008
Today we read about Coulomb's Law in the green textbook and worked on a few practice questions. We also handed in several textbook questions for marks, using Coulomb's Law to calculate the force between two charges. This took up the majority of the period and the class packed up waaay too early today!
May 15, 2008
It seems to me that my fellow physics friends forgot to scribe again so I'll do a good deed and grab myself some BONUS marks. First, we started our new unit about electric and magnetic fields. We described the similarities and difference about Newtons gravitational law and coulombs law. Second, we had to read pages 415-418 on the green book. Also we were assigned to hand in some questions which is due for tomorrow.
Monday - May 12, 2008
So I'll also be doing the scribe for May 12.
On Monday, we went over the questions in the green textbook; 17, 20, 21, 22, and 23 on the board. It took a while and so afterwards, near the end of class, we watched a short video about space.
On Monday, we went over the questions in the green textbook; 17, 20, 21, 22, and 23 on the board. It took a while and so afterwards, near the end of class, we watched a short video about space.
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