
#1
Feb2612, 04:30 PM

P: 123

Hi guys I have 3 question the first one I want to check if my answer is correct or not but the two other question I don't know how to approach them thanks.
1)The 500 gram mass carries a charge of –7 x 108 C. If the coefficient of friction between the mass and the surface is 0.20, and an electric field of magnitude 3 x 107 N/C is horizontal and directed to the right, which is being treated as the positive xdirection. What is the acceleration of the mass? Ff = force of friction; Ff = 0.2 * (500 * 10^3) = 10^1N; Fe = qE; Fe = (7 * 10^8) * (3 * 10^7) = 2.1N; Fnet = 2.1  10^1 = 2.2N; F = ma; 2.2 / (500 * 10^3) = a; > a = 4.4 m/s^2; I can't find that answer in the multiple question,so did I do something wrong here? 2)What is the magnitude and direction of the electric field generated by a 13 μC charge in a point 8.90 cm away from the charge? How can I know the electric field generated even though I am not given a point charge ? Thats what confusing me in this question. 3)A +2 μC charge is at the origin. A +5 μC charge is on the xaxis at +1 m. At what point between the charges is the electric field zero? I first calculated the electric field Electric field from F = kQ1 * Q2 / d^2. I got electric field of the 5μC = (9 * 10^4) / 5 and the electric field of the 2μC = (9 * 10^4) /2. I don't know how to proceed from here to get the distance of the zero electric field. This is not a homework or anything I am solving some electrostatic problems on my own. Thanks for the help in advance. 



#2
Feb2612, 06:08 PM

Mentor
P: 11,416





#3
Feb2612, 07:33 PM

P: 123

Thanks alot.
I still have problem for number 3 but I calculated the other problems as follows> I have calculated problem 1 to be fnet = 2.1 + 1 = 1.1; f = ma; a = 1.1/500 * 10^3; a = 2.2 m/s^2. For problem two its really easy E = (9 * 10^9)(13 * 10^6) / (8.9^2) = 1.48 * 10^7. I am still confused though for problem three I meant to say I first calculated the force then I went to see which distance will the force be zero but got nowhere from their. I tried to resolve it by assuming that point charge for the electric field of the 2 is 5 and vice versa and got the equation. E(due to 5) = (9 * 10^9)(5 * 10^6) / (1 + x)^2. E(due to 2) = (9 * 10^9)(2 * 10^6) / (1 + x)^2. Then I added E(due to 5) + E(due to 2) = 0; but got complex solution I think here my logic is flawed. I know I must did something wrong in this procedure. 



#4
Feb2612, 08:04 PM

Mentor
P: 11,416

Electrostatic problems helpIn these problems where you're dealing with quantities that involve the same multipliers and constants you can usually discard the common factors when ratios or equivalences are called for. So let the charge at the origin be +2, the charge at x=1 be +5. Between the charges the field from the +2 charge will point to the right while the field from the +5 charge will point to the left. So clearly there will be a point between them where the "right pointing" and "left pointing" will exactly cancel. If they cancel that means that their magnitudes will be the same. [tex] \frac{2}{x^2} = \frac{5}{(1  x)^2}[/tex] 



#5
Feb2612, 08:24 PM

P: 123

Yes I made my distance expression wrong. I thought the electric field of 2 due to point charge 5 should be the distance of 1 where they are far apart plus x,which is the distance the charges cancell. I followed same logic in my other Electrical field,but yes that is wrong since I am adding a distance of 1 that is why my answer was flawed.
Thanks. 


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