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b. Calculate the electric force that exists between two objects that are 0.500 m apart and carrycharges of 0.00450 C and 0.00240

Sagot :

Given

The two charges,

[tex]\begin{gathered} q_1=0.00450\text{ C} \\ q_2=0.00240\text{ C} \end{gathered}[/tex]

Distance between them,

[tex]r=0.5\text{ m}[/tex]

To find

The electric force

Explanation

We know the force is given by

[tex]F=\frac{kq_1q_2}{r^2}[/tex]

Putting the values,

[tex]\begin{gathered} F=9\times10^9\frac{0.00450\times0.00240}{(0.500)^2} \\ \Rightarrow F=388,800\text{ N} \end{gathered}[/tex]

Conclusion

The electric force is 388,800 N