Double checking charges and electrical field problemsby mr_coffee Tags: charges, checking, double, electrical, field 

#1
Aug3005, 07:17 PM

P: 1,629

Hello everyone, I've done a series of physics problems, but I have no way of seeing if they are right or wrong, So i'm going to copy and paste them, and then show you what I did to come up with the answer, I don't expect 1 person to double check all my problems, but if somtime when one or 2 of you are bored can you make sure I did them correctly? Thanks! If this isn't allowed, i'll delete the post immediatley! I Just want to make sure I get a good grade on this.
#2. A 5.0 C charge is 10 m from a 2.0 C charge. The electrostatic force is on the positive charge is: 9.0 108 N toward the negative charge F = k (5.0C)(2.0C)/(10m)^2 = 9.0x10^8, toward the negative charge because opposites attract. #4. Two small charged objects repel each other with a force F when separated by a distance d. If the charge on each object is reduced to onefourth of its original value and the distance between them is reduced to d/2 the force becomes: F = (kq^2)/(d/2) F = [(1/4)^2*2]/d = F/8 #5. What is the magnitude of a point charge that would create an electric field of 3.20 N/C at points 3.10 m away? E = 1/(4PIEo) (q/r^2); E(4PIEo)(r^2) = q (3.20)(4PI)(8.85x10^12)(3.10)^2 = q q = 3.42x10^9 #6. The electric field due to a uniform distribution of charge on a spherical shell is zero: Only inside the shell #7. In Figure 2224, two particles of charge q are arranged symmetrically about the y axis; each produces an electric field at point P on that axis. http://img293.imageshack.us/img293/5161/23221om.gif2224 (a) Are the magnitudes of the fields at P equal? yes (b) Is each electric field directed toward or away from the charge producing it? toward (c) Is the magnitude of the net electric field at P equal to the sum of the magnitudes of the two field vectors (is it equal to 2E)? nowouldn't it be the sum of y compoents of E1 and E2? (d) Do the x components of those two field vectors add or cancel? cancel (e) Do their y components add or cancel? add (f) Is the direction of the net field at P that of the canceling components or the adding components? adding components (g) What is the direction of the net field? toward negative y #9. Figure 2211 shows four situations in which charged particles are fixed in place on an axis. http://img76.imageshack.us/img76/510/22115ph.gif2211 In which situations is there a point to the left of the particles where an electron will be in equilibrium? (Select all that apply.) situation b visually it was the only one that made senes to me. #10. In Figure 2229 the electric field lines on the left have twice the separation as those on the right. http://img339.imageshack.us/img339/83/hrw722290cb.gif2229 (a) If the magnitude of the field at A is 60 N/C, what force acts on a proton at A? E = F/q; F = (60 N/C)(1.6x10^19); F = 9.6x10^8 N; (b) What is the magnitude of the field at B? If the magnitude of the field at B is exactly twice as wide, would it just be 120 N/C, because A is 60 N/C ? #11. A particle of charge of +3.40 106 C is 12.0 cm distant from a second particle of charge of 2.20 106 C. Calculate the magnitude of the electrostatic force between the particles. F = [K(q1)(q2)]/r^2; F = (9E9)(3.40E6)(2.20E6)/(.12)^2; F = 4.675N #12. The diagram shows two identical positive charges Q. The electric field at point P on the perpendicular bisector of the line joining them: http://img339.imageshack.us/img339/7912/f230211bk.jpgDiagram Downwards i drew the Efield lines and it looks like everyhtiung else cancels out but the y components downward. #13. Two protons (p1 and p2) and an eletron (e) lie on a straight line, as shown. The directions of the force of p2 on p1, the force of e on p1, and the total force on p1, respectively, are: http://img374.imageshack.us/img374/2714/f220330hu.jpgDiagram 2 to the left, to the right, to the left #14. In Figure 2126, particle 1 of charge +1.0 µC and particle 2 of charge 2.0 µC, are held at separation L = 13.0 cm on an x axis. If particle 3 of unknown charge q3 is to be located such that the net electrostatic force on it from particles 1 and 2 is zero, what must be the coordinates of particle 3? http://img291.imageshack.us/img291/8479/21262aw.gif2126 X = ?cm Y = ?cm This one i am lost on, any hints would be great!! #15. In Figure 2124a, particles 1 and 2 have charge 45.0 µC each and are held at separation distance d = 2.70 m. http://img291.imageshack.us/img291/9029/hrw721247yl.gif 2124a (a) What is the magnitude of the electrostatic force on particle 1 due to particle 2? F = [(9.9E9)(45.0E6)^2]/ (2.70m)^2 F= 2.75 N (b) In Figure 2124b, particle 3 of charge 45.0 µC is positioned so as to complete an equilateral triangle. What is the magnitude of the net electrostatic force on particle 1 due to particles 2 and 3? I drew a free body diagram of particle 1, and I'm alittle lost on what the angle between the two forces should be, I know 180 = sum of the angles of a triangle. So i'm thinking it should be 60. Then do i just find the resultant Force vector and thats the answer? Thanks. Thanks everyone! Just pick and choose! any help would be great! 



#2
Aug3005, 07:41 PM

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P: 11,154

#2 : correct
#4 : one mistake  you forgot to square the distance in the force equation 



#3
Aug3005, 07:55 PM

P: 1,629

Thanks so #4 should be F/4?




#4
Aug3005, 08:00 PM

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Double checking charges and electrical field problems#6 : correct (with an additional "trivial" solution being points at infinity) #8 : all correct 



#5
Aug3005, 08:10 PM

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P: 11,154

#9 : b is correct but there's another solution that you're missing.
Here's a useful way to solve this problem. For each case, consider two extreme points to the left of the charges : point A being very close to the first charge (distance to this charge is way smaller than the distance to the other charge or d1/d2 << 1), and point B being very far from both charges (or d1/d2 = 1, almost). At each point draw the two forces and their resultant. If the resultant changes direction between A and B, there must exist a point between A and B where the resultant = 0. 



#6
Aug3105, 04:00 AM

P: 7

i think for 14..it will be workable only by assuming the 3rd particle is a neutrally charged one and by trying to approximate the force experienced by this rticle at varous location such that the net force experienced by this particle is 0.....




#7
Sep105, 05:20 PM

P: 1,629

thanks for the replies guys, i had a question you said, #8 : all correct, did u mean #7? I never posted 8 i didn't think. Thanks!




#8
Sep105, 05:39 PM

P: 1,629





#9
Sep105, 08:10 PM

P: 166

Damn it! I just wrote answers to 10,11,12,13 and a long detailed answer to 14, and then just as I was about to submit it I clicked preview and internet explorer crashed. Oh well, give me a few hours and I'll come back and do it again.




#10
Sep105, 10:42 PM

P: 166

#10a correct
#10b when the field lines are further apart, it means that the field is weaker. #11 correct #12 To find the electric field consider what would happen to a positive test particle placed at point P. The force due to the first +Q would be to the right and slightly down. The force due to the second +Q would be to the right and slightly up. So the up and down components cancel and the resultant force is to the right. #13 correct #14 This one is more difficult. You need to resolve the forces for a point at an arbitrary position on the xaxis. You will have three different equations, one for to the left point 1, one for to the right of point 2, and one for between the two points. I'll do an example for you, for if you are considering to the right of point 2 (which is not where the q3 needs to be placed). So lets consider placing a particle of charge +q at a position x, where x > .13, and we will consider a force to the right to be positive and a force to the left to be negative. The force on q due to charge 1 is: [tex]kq\frac{1.0*10^6}{x^2}[/tex] The force on q due to charge 2 is: [tex]kq\frac{2.0*10^6}{(x.13)^2}[/tex] Notice that the force due to charge 1 is positive which means it pushes it to the right, and the force due to charge 2 is negative. For equilibrium we want these two forces to cancel out. [tex]\frac{1.0*10^6}{x^2}  \frac{2.0*10^6}{(x.13)^2} = 0[/tex] This is a quadratic equation, which has a solution at x=.054. But, this is outside the domain of our equation. When we set up this equation we assumed that charge 1 pushes q to the right, and charge 2 pushes q to the left. If q were in between charges 1 and 2, they would both push it to the right and we would need a different equation. So, there aren't any valid solutions to our equation, which means there aren't any points to the right of charge 2 with zero net electrostatic force. 



#11
Sep105, 10:49 PM

P: 166

#15a correct except that the constant should be 8.9E9 not 9.9E9
#15b Yep, that's the correct procedure. Give it a try and let me know what your answer is. 



#12
Sep405, 03:32 PM

P: 1,629

Thanks for the help kazaaa!! really nice of you. For 15a i got 2.5N For 15b I got 4.33N look good?




#13
Sep405, 06:36 PM

P: 166

No probs mr coffee. Yep, both of those are correct.




#14
Sep605, 06:39 PM

P: 1,629

Kazza, I used ur method to see if there was a point left of the 2nd charge and came out with this...
Hello everyone. I need to see if I did this right. the problem is this: Particle 1 of charge 1.0E6 C and particle 2 of charge 2.0E6 C are held at seperation L = 13.0cm on an x axis. If particle 3 of unkonwn charge q3 is to be located such that the net electrostatifc force on it from particles 1 and 2 is zero what must be the coordinates of particle 3? Here is my drawing and work: w00t So would the answer be X = 31.385 cm Y = 0 cm? does that seem right to you? 



#15
Sep705, 08:37 AM

P: 166

Yep, same answer I got, and your working is right, except remember it should be X=31.385 if I remember the original question correctly.




#16
Sep705, 10:57 AM

P: 1,629

thanks again!! i also thought it should be negative but in class i asked him and he said, no it should be positive, but if its the distance from the negative xaxis to the orgin, that would be negative i would think, or is distance always positive? or is this considered displacement?




#17
Sep705, 05:13 PM

P: 166

Yeah, distance is always positive, but the question asked what are the coordinates, so the coordinates are (31.385,0).



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