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What must the distance be between charges of +2.35 and −1.96 for the attractive force between them to be the same as that between charges of +4.06 and −2.11 separated by a distance of 2.26 pm?

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HomertheGenius
Force is:F = k * ( q1 * q2 ) / r²Therefore:k * ( 2.35 * 1.96 ) / r² = k * ( 4.06 * 2.11 ) / ( 2.26 )²   / : k4.606 / r² = 8.5666 / 5.10768.5666 r² = 4.606 * 5.1076r² = 23.5256 : 8.5666r² = 2.7462r = √2.7462Answer:r = 1.66 pm

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