Two point charges are separated by 10 cm. The attractive force between them is 26 N. If the two charges attracting each other have equal magnitude, what is the magnitude of each charge? (in coulombs)
2007-04-13 07:18:43 · 4 answers · asked by Dirck G 1 in Science & Mathematics ➔ Physics
f=q^2/r^2*4pi eo
26=9*10^9*q^2*100*100/100
26=9*10^11*q^2
q^2=26*10^-11/9
q=.53*10^-5
q=5.3*10^-6 C
2007-04-13 07:25:41 · answer #1 · answered by miinii 3 · 0⤊ 0⤋
F = Q*q / (4 * pi * eps0 * r^2)
Since Q and q are equal:
F = Q^2 / 4pi* epsilon0 * r^2
(r is in meters, epsilon0 = 8.85*10^-12 F/m )
26 N = Q^2 / 12.56 * 8.85*10^-12 F/m * 0.01 m^2
I think you can do the rest. Q will be in units of Coulombs.
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2007-04-13 07:30:17 · answer #2 · answered by tlbs101 7 · 0⤊ 0⤋
Use the inverse sq. regulation. The stress of attraction (or repulsion) between 2 factor costs is inversely proportinal to the sq. of their distance of seperation. (This stress is likewise immediately proportional to the made of their costs. yet this section is held consistent in this question) F1/F2 =d2^2/d1^2 10/F2 = 24^2/6^2 for this reason, F2 = 10/16 = 0.625 N.
2016-12-03 23:30:54 · answer #3 · answered by Anonymous · 0⤊ 0⤋
F = kq1xq2/d^2
26 = k q^2/(0.10)^2
Plug in the value for k and solve for q.
2007-04-13 07:25:45 · answer #4 · answered by Scott H 3 · 0⤊ 0⤋