Solve Physics Problem: EXAM I Physics Help

In summary, the Kirchhoff's current law states that the sum of all currents entering a junction must equal the sum of all currents leaving the junction. This can be applied to determine the current flow in a circuit.
  • #1
Alt+F4
306
0
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
 
Last edited by a moderator:
Physics news on Phys.org
  • #2
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:
  • #3
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
 
Last edited by a moderator:
  • #4
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.
 
Last edited by a moderator:
  • #5
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
 
  • #6
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.
 
  • #7
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
 
Last edited by a moderator:
  • #8
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
 
  • #9
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
 
Last edited by a moderator:
  • #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
 
Last edited:
  • #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/
 
Last edited by a moderator:
  • #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
 
Last edited by a moderator:
  • #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
 
Last edited by a moderator:
  • #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
 
Last edited by a moderator:
  • #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
 
Last edited by a moderator:
  • #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)
 
Last edited by a moderator:
  • #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
 
Last edited by a moderator:
  • #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 ([tex]\pi r^2[/tex]). 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.
 
Last edited by a moderator:
  • #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 ([tex]\pi r^2[/tex]). 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
 
Last edited by a moderator:
  • #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
 
Last edited by a moderator:
  • #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
 
Last edited by a moderator:
  • #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.
 
Last edited by a moderator:
  • #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.
 
Last edited by a moderator:
  • #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?
 
Last edited by a moderator:
  • #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
 
Last edited by a moderator:
  • #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 !...
 
Last edited by a moderator:
  • #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
 
Last edited by a moderator:
  • #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.
 

What is the best way to approach a physics problem on an exam?

The best way to approach a physics problem on an exam is to first carefully read and understand the question. Then, identify the relevant concepts and equations that can be applied. Next, draw a diagram or make a list of known and unknown variables. Finally, use the appropriate equations to solve for the unknown variable.

How can I improve my problem-solving skills in physics?

To improve your problem-solving skills in physics, it is important to practice solving a variety of problems. Start with simpler problems and gradually move on to more complex ones. Also, make sure to review and understand the underlying concepts and equations, rather than just memorizing them.

What should I do if I get stuck on a physics problem during an exam?

If you get stuck on a physics problem during an exam, take a deep breath and try not to panic. First, review the problem and make sure you understand what is being asked. Then, try to approach the problem from a different angle or use a different equation. If you are still stuck, move on to another problem and come back to it later if you have time.

How can I avoid making mistakes when solving physics problems on an exam?

To avoid making mistakes when solving physics problems on an exam, it is important to show all your work and double check your calculations. Make sure to use the correct units and pay attention to significant figures. Also, take your time and don't rush through the problems.

What resources can I use to study for my physics exam?

There are many resources available to help you study for your physics exam. These include textbooks, lecture notes, online tutorials, practice problems, and study groups. It is also helpful to attend review sessions or seek help from your instructor or a tutor if you are struggling with a particular concept.

Similar threads

Can I Use My Calculator for the Upcoming Physics Exam?
    • Introductory Physics Homework Help
2
Replies
66
Views
6K
Needing some help about time of flight of a launched ball/horiz range
    • Introductory Physics Homework Help
Replies
6
Views
2K
Not necessarily homework help but I do not understand Field lines
    • Introductory Physics Homework Help
Replies
2
Views
773
Sign Convention for Angular Acceleration in Rotational Motion
    • Introductory Physics Homework Help
Replies
15
Views
1K
Can someone please solve this physics problem?
    • Introductory Physics Homework Help
Replies
7
Views
1K
Can someone double check my answers to this practice exam?
    • Introductory Physics Homework Help
Replies
7
Views
1K
Calculator (root) problem (Exam tomorrow) HP 10s assistance
    • Introductory Physics Homework Help
Replies
11
Views
3K
Help with Physical Vector problem
    • Introductory Physics Homework Help
Replies
5
Views
1K
Calculating Work Done by a Constant Force on a Moving Body
    • Introductory Physics Homework Help
Replies
8
Views
6K
Physics olympiad problem -- struggling with polar coordinates
    • Introductory Physics Homework Help
2
Replies
37
Views
3K

Hot Threads

Recent Insights

Back
Top