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Two point charges are 3.00 cm apart. They are moved to a new separation of 2.00 cm. By what factor does the resulting mutual force between t
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Answer:
By a factor of 9/4
Explanation:
Applying Coulomb’s law,
F = kqq’/r²………………. Equation 1
Assuming q and q’ are the two point charges respectively.
Where k = coulomb’s constant, r = distance between the charges.
When the point charges are 3.0 m apart,
F = kqq’/3²
F = kqq’/9……………….. Equation 1
When they are moved to a new distance, 2.00 m
F’ = kqq’/2²
F’ = kqq’/4…………….. Equation 2.
Comparing equation 1 and equation 2.
F’ = 9F/4
Hence the resulting mutual force change by a factor of 9/4