Question

Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 33.55 N when they are separated by 55.75 cm. What is the magnitude of the charges in microCoulombs ?

Answer 1

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 ?

Step-by-step explanation

Let the charge be q.

We know the force of attraction is given by

`\begin{gathered} F=k(q^2)/(d^2) \\ \Rightarrow33.55=9*10^9*(q^2)/((0.5575)^2) \\ \Rightarrow q=3.39*(10)^(-5)C \\ \Rightarrow q=\pm33.9\mu C \end{gathered}`

Conclusion

The charges are:

`+33.9\mu C,-33.9\mu C`

The magnitude of equal charges are:

`\lvert{q}\rvert=33.9\mu C`