kingwinner Posted June 14, 2006 Report Share Posted June 14, 2006 For the folloing 6 questions, I got completely different (way off) answers from the ones given in the textbook. This is really driving my crazy. I checked my work many times but can't find out what I did wrong. If anyone is interested in physics problems and have the time, I hope you can help me check out whether the answers in the textbook are correct or not, so that I can know if I am correct. I really appreciate for your help! :shrug: I am wondering if I am calculating something seriously wrong so that I am getting way off answers for all these questions. I am afraid and frustrated now that I am losing confidence in this topic. :hihi: 1) A point charge of +3.8x10^-6 C is placed 0.20 m to the right of a charge of -2.0x10^-6 C. What is the force on a third charge of +2.3x10^-6 C if it is placed 1a) 0.10 m to the right of the second charge?[assuming the second charge is the charge on the right side(+3.8x10^-6C), I got an answer of 7.4N, but the answer is 7.9N ]1b) where would the third charge experience a net force of zero?[i got an answer of 0.53 m to the left of the -2.0x10^-6C charge, but the answer given is 1.2x10^-2 m [right of smaller charge] ] 2) 4 objects, each with a positive charge of 1.0x10^-6 C, are placed at the coreners of a 45-degree rhombus with sides of length 1.0m. Calculate the magnitude of the net force on each charge.[i got 2.2x10^-2 N and 1.9x10^-2 N respectively, but the answers provided are 8.4x10^-3 N and 1.9x10^-2 N respectively.] 3) 3 charges of +1.0x10^-4 C form an equilateral triangle with side length 40cm. What is the magnitude and direction of the electric force on each charge?[i got an answer of 9.7x10^2N[90 degrees from the line joining the other charges], but the answer provided is 8.9x10^2 N[90 degrees away from the line connecting other charges] ] Quote Link to comment Share on other sites More sharing options...

kingwinner Posted June 14, 2006 Author Report Share Posted June 14, 2006 4) 2 charges of +4.0x10^-6 C and +8.0x10^-6 C are placed 2.0m apart. What is the field strength halfway between them?[i got an answer of 3.6x10^4N/C toward the 4.0x10^-6C charge, but the answer given is 3.6x10^-4N/C toward the larger charge] 5) A point charge of 2.0x10^-6C experiences an electric force of 7.5N to the left. What force would be exerted on a -4.9x10^-5 C charge placed at the same spot?[My answer is 1.8x10^2N, but the answer given is 1.5x10^5N/C ] 6) [i got 3.2x10^5N/C [right], but the answer in the book is 1.8x10^5N/C ] Quote Link to comment Share on other sites More sharing options...

kingwinner Posted June 14, 2006 Author Report Share Posted June 14, 2006 I know you physictists can calculate these very quickly. :shrug: Could somebody help me with 1 or 2 questions? I still can't get the answers in the book! Thanks! Quote Link to comment Share on other sites More sharing options...

sebbysteiny Posted June 14, 2006 Report Share Posted June 14, 2006 1) this is a three stage process. Stage 1. Calculate the force on the charge from one of the charges. F1= q1q2/r^2 4*pi*E0 q1 = +3.8x10^-6 C, q2 = + 2.3x10^-6 C r = .2 + .1 = .3 F1 = 0.873N Stage 2. Calculate the force on the charge from the other charge. F2 = q3q2/r^2 4*pi* E0 q3 = -2.0x10^-6 C, q2 = + 2.3x10^-6 C, r = .1 F2 = -4.13N (ie to the left). Stage 3. Calcultate the resulting force F = F1 + F2 = -3.26N (ie to the left). That's how to work it out. However, I notice I have got the wrong answer. That is because I did not draw the charge picture correctly. This is partly because I don't have a clue what the question means by '0.1m to the right of the second charge' and partly because I didn't read it properly. Nevertheless, the simple easy method works. If the question were in 3 dimensions, it would still work. You just add the Force vectors together. Quote Link to comment Share on other sites More sharing options...

kingwinner Posted June 15, 2006 Author Report Share Posted June 15, 2006 for q1a, I think we should assume the second charge is the charge on the right side(+3.8x10^-6C, but I still don't get the answer. This set of questions is making me crazy... Quote Link to comment Share on other sites More sharing options...

IDMclean Posted June 15, 2006 Report Share Posted June 15, 2006 I don't know if it helps or not but the system presented is attractive in nature. Quote Link to comment Share on other sites More sharing options...

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