1. For the dipole in the figure , the charges are 2.60 separated by a distance o
ID: 2041758 • Letter: 1
Question
1. For the dipole in the figure , the charges are 2.60 separated by a distance of 3.90 and oriented so that = 35.0. If the electric field has a magnitude of 7400 , find the magnitude and direction (clockwise or counterclockwise) of the net torque on the dipole about the same axis as before.(answer in N*m)2. Each of two small spheres is positively charged; the combined charge totals 60.0 .
What is the charge on each sphere if the spheres are repelled with a force of magnitude 0.270 when placed 0.100 apart? (answer in C)
Explanation / Answer
QUESTION 2 Q1 + Q2 = 60nC. Moreover, the force of repulsion (meaning they carry the samecharge in terms of sign) is 0.27mN. Since the total charge ispositive, both have to be positively charged. So, 0.27 x 10^-3 = (9x10^9)(Q1)(Q2) / (R^2) Where r is the distance between them. Therefore, (Q1)(Q2) = (0.27 x 10^-3 x 0.1x0.1)/(9x10^9) =3x10^-16. Now, you have two equation in Q1 and Q2. In the equationabove, you can substitute Q2 = (60 x 10^9) - Q1 Thus, 60x10^9(Q1) - (Q1)^2 = 3x10^-16. Use your GDC to calculate the value of Q1 and Q2. You get twovalues 1 of which is Q1 and the other Q2. You can assign the valuesanyway you like since it does not say that Q1 is greater (orlesser) than Q2 ====================================== Let q and Q are the individual charges. Given that, q + Q = 60 nC = 6 x 10^-8 C. ------ (1) According to Coulomb's law, force between charges separated by a distance r is F = kqQ/r^2 where, k = 9 x 10^9 if the charges are placed in air. Given, F = 0.270 mN = 0.270 x 10^-3 N and r = 0.1 m = 10^-1 m Then, 0.270 x 10^-3 = 9 x 10^9 x qQ/(10^-1)^2 or qQ = 0.03 x 10^-14 = 3 x 10^-16 C^2 ----------- (2) Solving equations (1) and (2) gives q = 52.4 nC and Q = 7.6 nC or vice versa ============================= http://www.cramster.com/answers-jan-11/physics/small-spheres-positively-charged-combine_1104854.aspx?rec=0
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