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Two equal, but oppositely charged particles are attracted to each other electrically. The size of the force of attraction is 56.46 N when they are separated by 50.91 cm. What is the magnitude of the charges in microCoulombs?

Sagot :

Answer:

q = 40.3 microCoulombs

Explanation:

The force of attraction between two charges is given by the formula:

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

The force of attraction. F = 56.46 N

The separation, r = 50.91 cm

r = 50.91/100

r = 0.5091 m

The electrostatic constant is:

[tex]k=9\times10^9Nm^2C^{-2}[/tex]

Solve for the magnitude of charge q

[tex]\begin{gathered} F=\frac{kq^2}{r^2} \\ \\ 56.46=\frac{9\times10^9\times q^2}{0.5091^2} \\ \\ 56.46\times0.5091^2=9\times10^9\times q^2 \\ \\ q^2=\frac{56.46\times0.5091^2}{9\times10^9} \\ \\ q^2=1.63\times10^{-9} \\ \\ q=\sqrt{1.63\times10^{-9}} \\ \\ q=4.03\times10^{-5}C \\ \\ q=40.3\mu C \end{gathered}[/tex]

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