Question

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 1

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

`F=k(|q_1q_2|)/(r^2)`

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:

`k=8.99*10^9N(m^2)/(C^2)`

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

`F=(kq^2)/(r^2)`

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:

`\begin{gathered} q^2=(F)/(k)r^2 \\ \\ \Rightarrow q=\sqrt{(F)/(k)r^2}=r\sqrt{(F)/(k)} \end{gathered}`

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:

`\begin{gathered} q=r\sqrt{(F)/(k)} \\ \\ =(24.04*10^(-2)m)*\sqrt{(87.96N)/(8.99*10^9N(m^2)/(C^2))} \\ \\ =23.779...*10^(-6)C \\ \\ \approx23.78\mu C \end{gathered}`

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