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Suppose two equal charges of 0.65 C each are separated by a distance of 2.5 km in air. What is the magnitude of the force acting between them, in newtons?

Respuesta :

Answer:

Force between two equal charges will be 608.4 N

Explanation:

We have given charges [tex]q_1=0.65C\ and\ q_2=0.65C[/tex]

Distance between the charges = 2.5 km = 2500 m

According to coulombs law force between two charges is given by

[tex]F=\frac{1}{4\pi \varepsilon _0}\frac{q_1q_2}{r^2}=\frac{Kq_1q_2}{r^2}[/tex], here K is constant which value is [tex]9\times 10^9Nm^2/C^2[/tex]

So force [tex]F=\frac{9\times 10^9\times 0.65\times 0.65}{2500^2}=608.4N[/tex]

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