# Homework Help: EXAM I physics Help

1. Sep 21, 2006

### Alt+F4

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

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 dont understand why the signs are the way they are

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2. Sep 21, 2006

### nazzard

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

Last edited: Sep 21, 2006
3. Sep 21, 2006

### Alt+F4

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

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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4. Sep 21, 2006

### nazzard

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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5. Sep 21, 2006

### Alt+F4

ya so how do u know wat is leaving and entering, i have my circuit labeled but i dont see it

6. Sep 21, 2006

### nazzard

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.

7. Sep 22, 2006

### Alt+F4

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

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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8. Sep 22, 2006

### Alt+F4

ya i am still no seeing that cause i had a HW due and the equation was
I1 = I2+I3

9. Sep 23, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2w.pl?practice/exam1/su02 [Broken]

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. Sep 23, 2006

### big man

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.

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...

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. Sep 23, 2006

### Alt+F4

I thought stuff goes from postive to negative so -2 should be going dwon

Last edited: Sep 23, 2006
12. Sep 23, 2006

### big man

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. Sep 23, 2006

### Alt+F4

got it thanks

14. Sep 23, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp06 [Broken]

Question 5

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

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15. Sep 23, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp06 [Broken]

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. Sep 23, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa05 [Broken]

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. Sep 23, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp05 [Broken]

Question 13, i have no idea how u determine wat is in series and parallel with wat

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18. Sep 23, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp06 [Broken]

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. Sep 23, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/sp06 [Broken]

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. Sep 23, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa03 [Broken]

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

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21. Sep 24, 2006

### big man

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.

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22. Sep 24, 2006

### Alt+F4

Thank you sooo much

23. Sep 24, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2a.pl?practice/exam1/fa03 [Broken]

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

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24. Sep 24, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2w.pl?practice/exam1/su06 [Broken]
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. Sep 24, 2006

### Alt+F4

http://online.physics.uiuc.edu/cgi/courses/shell/phys102/fall06/prep2w.pl?practice/exam1/su06 [Broken]
Question 24

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

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