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Oksana_A [137]
4 years ago
9

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?
Physics
1 answer:
quester [9]4 years ago
4 0

Answer:

F = - 1,598 10⁻³ N

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

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Earth has a magnetic dipole moment of 8.0 × 1022 J/T. (a) What current would have to be produced in a single turn of wire extend
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Answer:

a

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b

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c

No this arrangement can not be used to cancel out earths magnetic field at points on the Earth's surface .this because on the earth surface it shifts from its behavior as a magnetic dipole

Explanation:

    From the question we are told that

             The magnetic moment of earth is M = 8.0*10 ^{22} J/T

               The radius of earth generally has a value of R = 6378 *10^3 m

Magnetic moment is mathematically given as

                    M = IA

A is the area of the of the earth(assumption that the earth is circular ) and this evaluated as  

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Now making I the subject in the above formula

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                     = \frac{8.0^10^{22}}{\pi (6378 *10^{3})^2}

                     = 6.26 *10^8 A

                   

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