Given
Two equal, but oppositely charged particles are attracted to each other electrically.
Force of attraction is F=33.55 N
Distance between them, d=55.75 cm=0.5575 m
To find
What is the magnitude of the charges in microCoulombs ?
Explanation
Let the charge be q.
We know the force of attraction is given by
[tex]\begin{gathered} F=k\frac{q^2}{d^2} \\ \Rightarrow33.55=9\times10^9\times\frac{q^2}{(0.5575)^2} \\ \Rightarrow q=3.39\times(10)^{-5}C \\ \Rightarrow q=\pm33.9\mu C \end{gathered}[/tex]Conclusion
The charges are:
[tex]+33.9\mu C,-33.9\mu C[/tex]The magnitude of equal charges are:
[tex]\lvert{q}\rvert=33.9\mu C[/tex]