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Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 87.96 N when they are separated by 24.04 cm. What is the magnitude of the charges in microCoulombs ?

Answer :

The magnitude of the electric force between two charged particles is given by Coulomb's Law:

[tex]F=k\frac{|q_1q_2|}{r^2}[/tex]

Where q_1 and q_2 are the charges of the particles, r is the distance between the charged particles and k is the Coulomb's Constant:

[tex]k=8.99\times10^9N\frac{m^2}{C^2}[/tex]

Since the magnitude of both charges is the same, the equation becomes:

[tex]F=\frac{kq^2}{r^2}[/tex]

The force and the distance between the particles are given, the value of k is known and the charge q is unknown. Isolate q from the equation:

[tex]\begin{gathered} q^2=\frac{F}{k}r^2 \\ \\ \Rightarrow q=\sqrt{\frac{F}{k}r^2}=r\sqrt{\frac{F}{k}} \end{gathered}[/tex]

Replace the values of r=24.04*10^-2m, F=87.96N as well as the value of k to find the magnitude of the charges:

[tex]\begin{gathered} q=r\sqrt{\frac{F}{k}} \\ \\ =(24.04\times10^{-2}m)\times\sqrt{\frac{87.96N}{8.99\times10^9N\frac{m^2}{C^2}}} \\ \\ =23.779...\times10^{-6}C \\ \\ \approx23.78\mu C \end{gathered}[/tex]

Therefore, the magnitude of the charges in microCoulombs is: 23.78μC.

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