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sammy [17]
3 years ago
9

Two small identical metal spheres, A and B, on insulated stands, are each given a charge of +2.0 x 10-5 coulomb. The distance be

tween the spheres is 2.0 x 10-1 meter. Calculate the magnitude of the electrostatic force that the charge on sphere A exerts on the charge on sphere B.
a) 90 N
b) 18 N
c) 66 N
d) 130 N
Physics
1 answer:
SpyIntel [72]3 years ago
5 0

Answer:

a) 90 N

Explanation:

Coulomb's law: states that the electric force of attraction or repulsion between two charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them.

It is represented mathematically as

F = 1/4πε₀(q₁q₂)/d²......................... Equation 1

Where F = Electrostatic force, q₁ = charge on the first metal sphere, q₂ = charge on the second metal sphere, d = distance between the spheres, 1/4πε₀ = proportionality constant.

<em>Given: q₁ = +2.0×10⁻⁵ C, q₂ = 2.0×10⁻⁵, d = 2.0×10⁻¹ m</em>

<em>Constant : 1/4πε₀ = 9×10⁹Nm²/C²</em>

<em>Substituting these values into equation 1,</em>

<em>F = (9×10⁹×2.0×10⁻⁵×2.0×10⁻⁵)/(2.0×10⁻¹)²</em>

<em>F = (9×4×10⁻¹)/4×10⁻²</em>

<em>F = 90 N</em>

Therefore the electrostatic force is = 90 N

The right option is a) 90 N

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         f = \frac{1}{2 \pi }  \sqrt{ \frac{ 1000 \ \pi  \ 0.03^2 \ 9.8 }{0.025}  }

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