PROBLEMAS DE LEY DE COULOMB.
1)
f = k (-8x10-6c) (-6x10-6c)
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( 5x10-3m)2
f = k (48x10-12c)
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25x10-6 m2
f = k (1.92x10-12+6 c)N
f = k (1.92x10-6) N
f = k (0.00000192)N
2)
f = k ( 2x10-6c) (-6x10-6c)
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(0.02m)
f = k (2x10-6c) (-6x10-6c) = f = k (-12x10-12c)
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(0.02m)2 4X10-4m
f = k (-12x10-12c)
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4x10-4m
f = k (-3x10-12+4c)N
f = k ( -3x10-8)N
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