Asked by Anonymous
How much energy is need to place four positive charges, each of magnitude +5.0 mC, at the vertices of a square of side 2.5 cm?
Answers
Answered by
drwls
Compute the potential energy of the configuration, assuming that the charges are brought in from infinity one at a time. The will be the energy required. Let each corner charge be q and the square side length a. brining in the first charge requires no work. The second charge is placed a distance a from the first charge. The third charge gets placed a distance a from one charge and sqrt2*a from the other charge. Then bring in the fourth charge. You get the idea.
W = k[q/a + q/a + q/(a*sqrt2) + q/a + q/a + q/(a*sqrt2)]
= (k/a)[4 + sqrt2]
k is the Coulomb's law constant. Make sure q is in coulombs and a is in meters to get the answer in joules.
W = k[q/a + q/a + q/(a*sqrt2) + q/a + q/a + q/(a*sqrt2)]
= (k/a)[4 + sqrt2]
k is the Coulomb's law constant. Make sure q is in coulombs and a is in meters to get the answer in joules.
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