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Need some help in physics - I missed a class yesterday

N

neonerd

Diamond Member
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 😛, 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 😕

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?
 
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 😛, 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 😕

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.
 
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
 
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
 
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
 
Don't get your positives and negatives mixed up. Those are easy points to lose. You need to declare your positive reference point.
 
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? 😕
 
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