As shown in the figure on the left, a positive test charge
placed exactly halfway between a ball of charge +Q and a
second ball of unknown charge experiences a net force directed
right. When the second ball is removed from the situation, as in
the figure on the right, the force experienced by the test charge
increases by a factor of 1.60. What is the sign and magnitude of
the charge on the second ball?
Please explain how you got the answer!
Homework Answers
here,
the initial force on the test charge , F1 = K * Q * q /(r^2) - K * Q2 * q /r^2
F1 = K * q /r^2 * ( Q - Q2) . .....(1)
when the second charge is removed
the force on the test charge , F2 = 1.6 * F1
K * Q * q /r^2 = 1.6 * F1 ....(2)
from (1) and (2)
1/1.6 = (Q - Q2) /Q
Q = 1.6 * Q - 1.6 * Q2
Q2 = 0.375 Q
the charge on second ball is 0.375 Q
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