Answered

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An object carries a +15.5 µC charge. It is 0.525 m from a -7.25 µC charge. What is the magnitude of the electric force on the object?

Sagot :

onyebuchinnaji

Answer:

the force of attraction between the two charges is 3.55 N.

Explanation:

Given;

first charge carried by the object, q₁ = 15.5 µC

second charge carried by the q₂ = -7.25 µC

distance between the two charges, r = 0.525 m

The force of attraction between the two charges is calculated as;

[tex]F = \frac{kq_1q_2}{r^2} \\\\F = \frac{(9\times 10^9)(15\times 10^{-6})(7.25\times 10^{-6})}{(0.525)^2} \\\\F = 3.55 \ N[/tex]

Therefore, the force of attraction between the two charges is 3.55 N.

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