Need some help in physics - I missed a class yesterday

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neonerd

Diamond Member
Apr 24, 2003
8,746
1
0
I need somebody to explain to me specifically "Forces of changed particles & Coulomb's Law".

Basically, I know that 8.99 x 10^-9 is a constant used in the beginning of the equation, and then after you put (q1 and q2), which i have no clue what it means, but i know it's the numbers that correspond to the particles :p, and divide everything by distance ^2

I saw some people use 1.6 x 10 ^-19 for some problems, and the teacher said it was a constant too...I just have no clue when to use this constant, and when not to use it. It was used in some problems, but not in others :confused:

Here's an example of one of the problems we have to do:

Particles of charge +70, +48, and -80 µC are placed in a line. The center one is 0.35 m from each of the others. Calculate the force on each charge due to the other two.

Diagram:

70µC...................48µC..............-80µC
O_____________O_____________O
...........0.35m.................0.35m

I began to understand it with two particles, but 3 completely lost me. I imagine 70, 48, and -80, would be q1, q2, and q3...d would be 0.35m. Would this be a problem where I use 1.6*10^-19?
 
P

PrinceofWands

Lifer
May 16, 2000
13,522
0
0
Originally posted by: neonerd
I need somebody to explain to me specifically "Forces of changed particles & Coulomb's Law".

Basically, I know that 8.99 x 10^-9 is a constant used in the beginning of the equation, and then after you put (q1 and q2), which i have no clue what it means, but i know it's the numbers that correspond to the particles :p, and divide everything by distance ^2

I saw some people use 1.6 x 10 ^-19 for some problems, and the teacher said it was a constant too...I just have no clue when to use this constant, and when not to use it. It was used in some problems, but not in others :confused:

Here's an example of one of the problems we have to do:

Particles of charge +70, +48, and -80 µC are placed in a line. The center one is 0.35 m from each of the others. Calculate the force on each charge due to the other two.

Diagram:

70µC...................48µC..............-80µC
O_____________O_____________O
...........0.35m.................0.35m

I began to understand it with two particles, but 3 completely lost me. I imagine 70, 48, and -80, would be q1, q2, and q3...d would be 0.35m. Would this be a problem where I use 1.6*10^-19?
Click to expand...

Couldn't tell you...1.6x10^-19 coulomb is the constant of particle charge (the charge of a single proton, while -1.6x10^-19 is the charge of an electron). Other than that though you're out of my league.
 
S

simms

Diamond Member
Sep 21, 2001
8,211
0
0
mathworld.wolfram.com may help you out.

It's one freaking lecture. Geez.
 
N

neonerd

Diamond Member
Apr 24, 2003
8,746
1
0
Originally posted by: simms
mathworld.wolfram.com may help you out.

It's one freaking lecture. Geez.
Click to expand...

one lecture, with graded problems attached right to it ;)
 
C

Cawchy87

Diamond Member
Mar 8, 2004
5,104
2
81
I just did that in grade 12 physics. I can give you a scan tomorrow to mabye give you a base to start from.
 
N

neonerd

Diamond Member
Apr 24, 2003
8,746
1
0
My answer for some reason, comes out completely different from what the book says. Could somebody please do out the work for the first particle?

Here is my work:

F = (8.99*10^-9) (48*10^-6) (70*10^-6) / 0.35^2

which comes out to be: 2.46*10^-16

then F = (8.99*10^-9) (70*10^-6) (-80*10^-6) / 0.7^2

which comes out to be: -3.48*10^-16

And the answer I get it: 1.4*10^-16

Answer in the book says: -1.4*10^2N
 
Buttzilla

Buttzilla

Platinum Member
Oct 12, 2000
2,676
1
81
the question is talking about charges in relation to each other....

so if you have 3 particls: 1, 2, 3

and each of them have a charge and some distance from each other and you want to find charge 2 in relation to charge 1 and 3 (which is what i think your trying to ask) thats when you use the equation

F of charge 1 acting on charge 2: which equals F = ke (q1)(q2) / r^2

F of charge 3 acting on charge 2: F = ke (q3)(q2) / r^2
 
N

neonerd

Diamond Member
Apr 24, 2003
8,746
1
0
Originally posted by: Buttzilla
the question is talking about charges in relation to each other....

so if you have 3 particls: 1, 2, 3

and each of them have a charge and some distance from each other and you want to find charge 2 in relation to charge 1 and 3 (which is what i think your trying to ask) thats when you use the equation

F of charge 1 acting on charge 2: which equals F = ke (q1)(q2) / r^2

F of charge 3 acting on charge 2: F = ke (q3)(q2) / r^2
Click to expand...

I did that, trying to find charge 2 & 3 acting on charge one....i get 1.4*10^-16, whereas the answer should be -1.4*10^2

I've got part of it there....the 1.4, but I am completely oblivious how the book answer and my answer came out so different
 
T

TLfromAI

Senior member
Jun 22, 2002
379
0
0
Originally posted by: Cawchy87
I just did that in grade 12 physics. I can give you a scan tomorrow to mabye give you a base to start from.
Click to expand...


YOU GOT SERVED BEEATCH!!
 
TuxDave

TuxDave

Lifer
Oct 8, 2002
10,571
3
71
Don't get your positives and negatives mixed up. Those are easy points to lose. You need to declare your positive reference point.
 
N

neonerd

Diamond Member
Apr 24, 2003
8,746
1
0
Originally posted by: TuxDave
Don't get your positives and negatives mixed up. Those are easy points to lose. You need to declare your positive reference point.
Click to expand...

hm, asnwer is negative

i came out with -102 and 246 for the two forces...now i need to find the operation to make them equal -144

-102+246 = +144
-102-246 = -348
246+(-102) = +144
246-(-102) = 348

now is there really no way I can get -144 unless it's 102 and -246, or am is it getting a little late, and time for me to put down the crack pipe and get to bed? :confused:
 
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