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Two very small spheres are initially neutral and separated by a distance of 0.30 m. Suppose that 2.50 1013 electrons are removed from one sphere and placed on the other. (a) What is the magnitude of the electrostatic force that acts on each sphere?

User Netic
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1 Answer

4 votes

Answer:

F = - 1,598 10⁻³ N

Step-by-step explanation:

Electic strength is given by Coulomb's law

F = k q₁ q₂ / r²

Where k is the Coulomb constant that is worth 8.99 10⁸ N m²/C², q₁ and q₂ are the charges and r is the distance that separates the electric charges

In this case the charge of the two spheres is the same and of a different sign since when you remove the charge of a sphere that was initially neutral, it is left with that charge removed but of the opposite sign

q₁ = q₂ = 2.50 10¹³ electrons = 2.50 10¹³ 1.6 10⁻¹⁹

q₀ = 4.0 10⁻⁶ C

Let's calculate

F = - 8.99 10⁸ (4.0 10⁻⁶)² / 0.30²

F = - 1,598 10⁻³ N

User Apurv Agarwal
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