Answer:
Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere
Explanation:
The formula to be used here is
Q = It
where Q is the quantity of electricity and it is measured coulombs (C); 2.8 × 10⁻⁸ C or 0.000000028 C
I is current and it is measured in ampere (amps or A); unknown
t is time and it is measured in seconds (s); 0.05 s
Since, average current is what is unknown
I =Q/t
I = 0.000000028/0.05
I = 5.6 × 10⁻⁷ A
Average current produced by the repeated transfer of charge is 5.6 × 10⁻⁷ ampere
Answer:
A. F=107.6nN
B. Repulsive
Explanation:
According to coulombs law, the force between two charges is express as
F=(Kq1q2) /r^2
If the charges are of similar charge the force will be repulsive and if they are dislike charges, force will be attractive.
Note the constant K has a value 9*10^9
Hence for a charge q1=7.10nC=7.10*10^-9, q2=4.42*10^-9 and the distance r=1.62m
If we substitute values we have
F=[(9×10^9) ×(7.10×10^-9) ×(4.42×10^-9)] /(1.62^2)
F=(282.4×10^-9)/2.6244
F=107.6×10^-9N
F=107.6nN
B. Since the charges are both positive, the force is repulsive
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