Solve Physics Problem: EXAM I Physics Help

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The discussion revolves around solving physics problems related to forces and circuit analysis. Participants seek clarification on the signs of forces acting on charges, particularly why certain components are negative due to repulsion between like charges. Additionally, there is a focus on applying Kirchhoff's laws to analyze current flow in circuits, with discussions on identifying entering and leaving currents at junctions. Participants also address misconceptions about series and parallel configurations in circuits and the effects of dielectrics on capacitance and charge. Overall, the thread emphasizes understanding fundamental principles in physics to solve exam problems effectively.
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http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa00


Question 7


i know that the equation is



Fx = (-)(K)(12)(6) / (4^2) + K (12)^2 / (Square root of 32)) * Cos theta

Fy = K (12)(6) / 4^2 - K(12)^2 / (square root of 32)) * Sin theta




I just need to for somone to explain how come for the first variable is negative. I don't understand why the signs are the way they are
 
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Hello Alt+F4,

assume the x-axis "pointing" to the right and the y-axis to the top.

If you want to find the x-component Fx of the resulting force F on Q1 you'll find:
  • Q2 does not contribute
  • Q4 and Q1 will repel each other since they both are positive charges forcing Q1 in the "negative" x direction, hence the (-) sign
  • Q3 and Q1 will attract each other forcing Q1 in the positive x direction (and negative y direction, but that will be covered by Fy), hence the (+) sign

You can determine the correct signs for the y-component of the resulting force on the same way.

Regards,

nazzard
 
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http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa00

Question 23


I understand how to do 22. Answer is +E1 - I1R1 + I3R3 - E3 = 0

Now any trick to getting 23
 
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Alt+F4 said:
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa00

Now any trick to getting 23

No trick. Just a straight forward application of Kirchhoff's current law: sum of all currents entering the junction equals the sum of all currents leaving the junction.
 
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nazzard said:
No trick. Just a straight forward application of Kirchhoff's current law: sum of all currents entering the junction equals the sum of all currents leaving the junction.
ya so how do u know wat is leaving and entering, i have my circuit labeled but i don't see it
 
Alt+F4 said:
ya so how do u know wat is leaving and entering, i have my circuit labeled but i don't see it

The small triangles/arrows next to I1, I2 and I3 represent the (randomly!) chosen directions. I say randomly, because if you actually try to analyze a circuit you will know the correct "directions" for the currents after you've done all calculations. It will be represented by the sign of your result.
 
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa00

Question 6

Why am i doing E = 2 *(( (9*10^9)(12*10^-9) / (R^2)))


i thought it should have been 144 *10^-9 instead of 12
 
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nazzard said:
The small triangles/arrows next to I1, I2 and I3 represent the (randomly!) chosen directions. I say randomly, because if you actually try to analyze a circuit you will know the correct "directions" for the currents after you've done all calculations. It will be represented by the sign of your result.
ya i am still no seeing that cause i had a HW due and the equation was
I1 = I2+I3
 
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2w.pl?practice/exam1/su02

Question 13

All right so for the X direction the +2UC should be attracting to the -2UC forcing it in the + X-Direction since postive goes to negative
 
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  • #10
Now any trick to getting 23

As nazzard said there is no trick here. The current entering a junction must equal the current leaving a junction. So as you can see I1 and I3 are entering the junction. So that means that the sum of those currents must equal the current leaving the junction I2.

Question 6

Why am i doing E = 2 *(( (9*10^9)(12*10^-9) / (R^2)))i thought it should have been 144 *10^-9 instead of 12

You don't need to really calculate the force. Just draw the force vectors. You know that the two positive charges are going to repel the 12 uC charge and that in the horizontal and vertical directions. So add those two vectors and you will get a force that is along the diagonal. Now for the -12 uC charge you know it will attract the 12 uC charge along the diagonal. So the obvious choice is...

Question 13

All right so for the X direction the +2UC should be attracting to the -2UC forcing it in the + X-Direction since postive goes to negative

No it won't be forcing it in the positive x direction. -ve x-direction is the left and the +ve x-direction is on the right. Since the +2 uC charge will be attracting the -2 uC charge towards it, it will be forcing it left. Then you need to add the effect of the -4 uC charge on the -2 uC charge to find the overall direction.
 
  • #11
No it won't be forcing it in the positive x direction. -ve x-direction is the left and the +ve x-direction is on the right. Since the +2 uC charge will be attracting the -2 uC charge towards it, it will be forcing it left. Then you need to add the effect of the -4 uC charge on the -2 uC charge to find the overall direction.


I thought stuff goes from postive to negative so -2 should be going dwon
 
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  • #12
Yes the -2 uC charge will be forced downwards due to the repulsion force from the -4 uC charge and then it will be moved in the positive y and negative x direction due (ie diagonally) to the +2 uC charge. However because the repulsion force is much greater the attractive force from the +2 uC charge the -2 uC charge will still be forced downwards and to the left.
 
  • #13
big man said:
Yes the -2 uC charge will be forced downwards due to the repulsion force from the -4 uC charge and then it will be moved in the positive y and negative x direction due (ie diagonally) to the +2 uC charge. However because the repulsion force is much greater the attractive force from the +2 uC charge the -2 uC charge will still be forced downwards and to the left.
got it thanks
 
  • #14
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp06

Question 5


All right so i was thinkin it is going to be I1 = I2+I3 is that correct/
 
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  • #15
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp06

Question 11

I know C1+C2 = .6666 since they are in series and that Q=CV


I know C how do i find Q
 
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  • #16
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa05


Question 18, I still have no idea why it is the way it is i just looked the answer and it turns out to be I1-I2 = I3
 
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  • #17
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp05

Question 13, i have no idea how u determine wat is in series and parallel with wat
 
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  • #18
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp06

Question 5

Do u have to have 3 equations and u would then need to find I1 and I2 to get I3
 
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  • #19
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp06

last question of the day
17.

The Formula is R = P L/A so wat i did was (.025*10^-6)(30) / ( 100*10^-6 *30)
 
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  • #20
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa03

Somone guide me for 21, they all look in series to me
 
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  • #21
http://online.physics.uiuc.edu/cgi/c...ice/exam1/sp06

Question 5


All right so i was thinkin it is going to be I1 = I2+I3 is that correct/

Yes that's right. Now you just need two more equations (making three in total) for your three unknowns and you can solve for I3.

The two other equations are going to be from Kirchoff's loop rule.


Question 11

I know C1+C2 = .6666 since they are in series and that Q=CV


I know C how do i find Q

Yeah that's right as well. The capacitors in series are going to add like resistors in parallel. The one thing with capacitors in series is that the charge is going to be common to all the capacitors. The only thing that changes is the voltage. So after a long time the capacitors are fully charged. Since C2 is twice as large as C1, from the Q=CV equation we see that the voltage across C2 must be half of C1 to produce the same charge.

Question 18, I still have no idea why it is the way it is i just looked the answer and it turns out to be I1-I2 = I3

OK look at the diagram for Q18. You see the arrow of I1 follow the circuit around in that direction until you reach the juntion that is after R1. As you can see when I1 reaches that junction it splits into two different direction. Part of I1 sent down the path of I2 and part of I1 is sent down the path of I3. So this means that I1 is made up of I2 and I3, hence I1=I2+I3. Re-arranging you get I1-I2=I3.

last question of the day
17.

The Formula is R = P L/A so wat i did was (.025*10^-6)(30) / ( 100*10^-6 *30)

You have the right formula here, but you've mad an error. The A refers to the cross-sectional area of the wire. The cross-sectional area is going to be the area of a circle (\pi r^2). Remember you are given the diameter so you need to use the radius, which is half that value. You will get the right answer if you do this.

Somone guide me for 21, they all look in series to me

C2 and C3 are obviously in parallel so you find the equivalent capacitance of those two first. Then you now see the remaining three capacitances are in series.
 
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  • #22
big man said:
Yes that's right. Now you just need two more equations (making three in total) for your three unknowns and you can solve for I3.

The two other equations are going to be from Kirchoff's loop rule.




Yeah that's right as well. The capacitors in series are going to add like resistors in parallel. The one thing with capacitors in series is that the charge is going to be common to all the capacitors. The only thing that changes is the voltage. So after a long time the capacitors are fully charged. Since C2 is twice as large as C1, from the Q=CV equation we see that the voltage across C2 must be half of C1 to produce the same charge.



OK look at the diagram for Q18. You see the arrow of I1 follow the circuit around in that direction until you reach the juntion that is after R1. As you can see when I1 reaches that junction it splits into two different direction. Part of I1 sent down the path of I2 and part of I1 is sent down the path of I3. So this means that I1 is made up of I2 and I3, hence I1=I2+I3. Re-arranging you get I1-I2=I3.



You have the right formula here, but you've mad an error. The A refers to the cross-sectional area of the wire. The cross-sectional area is going to be the area of a circle (\pi r^2). Remember you are given the diameter so you need to use the radius, which is half that value. You will get the right answer if you do this.



C2 and C3 are obviously in parallel so you find the equivalent capacitance of those two first. Then you now see the remaining three capacitances are in series.
Thank you sooo much
 
  • #23
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa03

Question 21 the one about the Resistors, i think you looked at 13. Thanks
 
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  • #24
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2w.pl?practice/exam1/su06
Question 22.

I thought that it should be Qi < Qf since any time u put in a dilelectric V has to stay the same, Capacitance Increases and therefore Q has to increase
 
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  • #25
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2w.pl?practice/exam1/su06
Question 24

I1 and I2 should meet so i thought it would be I1+I2 = I3, i guess not
 
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  • #26
http://online.physics.uiuc.edu/cgi/c...ice/exam1/fa03

Question 21 the one about the Resistors, i think you looked at 13. Thanks

Ok this is a wheatstone bridge arrangement. Anyway, we have a current flowing left and we follow that around until we reach the resistor arrangemnet. Now you can see that there are two different directions that the current can go in. This means that the top branch is in parallel with the bottom branch. And if we choose one direction (say the top branch) we see that the current can only follow the one path until it reaches the other side. This obviously means that the resistors on the top branch are in ...

http://online.physics.uiuc.edu/cgi/c...ice/exam1/su06
Question 22.

I thought that it should be Qi < Qf since any time u put in a dilelectric V has to stay the same, Capacitance Increases and therefore Q has to increase

Nope. Do you remember the relationship between separation distance of the plates of a capacitor and its capacitance? Do you also remember how that relationship changes when you add a dielectric?

Alt+F4 http://online.physics.uiuc.edu/cgi/c...ice/exam1/su06
Question 24

I1 and I2 should meet so i thought it would be I1+I2 = I3, i guess not

Yeah that would be the case if I3 and the arrow were on the direct opposite side of that circuit diagram. But if you follow I3 around in that direction you see that it points in the opposite direction to I1. So all the currents are enetering the junction and you don't have any leaving.

Try creating equations using kirchhoffs loop rule for the left hand loop and the right hand loop.
 
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  • #27
All right, i think this should be last question

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa02

27 and 28. So for 27, 5 of the postive charges cancel out with each other, so leaving 4 and one negative but how come the answer is 2kq/r^2 and for 28 I am guessing since 11 postives + 1 negatvie it will be 10.
 
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  • #28
All right, i think this should be last question

http://online.physics.uiuc.edu/cgi/c...ice/exam1/fa02

27 and 28. So for 27, 5 of the postive charges cancel out with each other, so leaving 4 and one negative but how come the answer is 2kq/r^2 and for 28 I am guessing since 11 postives + 1 negatvie it will be 10.

haha OK fair enough.

Why do only five of the +ve point charges cancel? Draw lines connecting opposite point charges. You will see that due to the symmetry all the charges will cancel except for two charges. Those two will be the -ve one and the +ve one opposite. They will essentially add up to get 2kq/r^2.

Yep you're right about Q28. V= kq/r and if you have multiple point charges you just sum the potentials.

By the way did you solve the capacitance one with the dielectric above alright?
 
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  • #29
big man said:
haha OK fair enough.

Why do only five of the +ve point charges cancel? Draw lines connecting opposite point charges. You will see that due to the symmetry all the charges will cancel except for two charges. Those two will be the -ve one and the +ve one opposite. They will essentially add up to get 2kq/r^2.

Yep you're right about Q28. V= kq/r and if you have multiple point charges you just sum the potentials.

By the way did you solve the capacitance one with the dielectric above alright?
yep thanks
 
  • #30
Maybe one more,
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp04


Question 10
 
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  • #31
Alt+F4 said:
Maybe one more,
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp04


Question 10
help !...
 
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  • #32
http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2w.pl?practice/exam1/su05

Qustion 22 , how come it is ln(2)*4*2*2 = 11.09 Sec i thought the resistor that i would plugged in would have been the EQuivalent of them all
 
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  • #33
help !...question 10

Well this one has a straightforward association between the electric field between the plates, the potential difference and the distance. V=Ed

http://online.cctt.org/physicslab/content/PhyAPB/lessonnotes/parallelplates/parallel_plates.asp

If your book doesn't have any theory on this then just go to the above site or better yet you can search hyperphysics for more info.

Qustion 22 , how come it is ln(2)*4*2*2 = 11.09 Sec i thought the resistor that i would plugged in would have been the EQuivalent of them all

Well like you were thinking the capacitor is discharging through the resistors. However, it won't be discharging through R1. It will only discharge through R2 and R3. This is because the path along R3 is the path of least resistance. The open switch is like having an infinite impedance so current will not flow down a path of infinite impedance...instead it will flow through the path marked by R2 and R3. This means R2 and R3 are in series, which makes an equivalent resistance of 8 ohms. This is why you get 11 microseconds.
 
  • #34
big man said:
Well this one has a straightforward association between the electric field between the plates, the potential difference and the distance. V=Ed

http://online.cctt.org/physicslab/content/PhyAPB/lessonnotes/parallelplates/parallel_plates.asp

If your book doesn't have any theory on this then just go to the above site or better yet you can search hyperphysics for more info.
Well like you were thinking the capacitor is discharging through the resistors. However, it won't be discharging through R1. It will only discharge through R2 and R3. This is because the path along R3 is the path of least resistance. The open switch is like having an infinite impedance so current will not flow down a path of infinite impedance...instead it will flow through the path marked by R2 and R3. This means R2 and R3 are in series, which makes an equivalent resistance of 8 ohms. This is why you get 11 microseconds.
we have no book, basically practice exams and lectures that are meanigless
 
  • #35
Like I said hyperphysics is a pretty good site at explaining most physics concepts so you might want to bookmark that site!

Also download emule plus http://sourceforge.net/project/showfiles.php?group_id=71866

Do a search for 'Physics for scientists and engineers'. It is a really good book that explains everything well. It's what I used in my first year.

The authors are Serway, Beichner and Jewett.
 

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