What must be the distance between point charge 91 = 20 MC and point charge 02 -45,0 AC for the electrostatic force between them
1 answer:
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Answer:
Explanation:
Given
,
,
,
The tension of the spring is
The force in the spring is equal to centripetal force so
But Fc is also
Fc=KxΔr
Replacing
total distance is
Answer:
q = 3.6 10⁵ C
Explanation:
To solve this exercise, let's use one of the consequences of Gauss's law, that all the charge on a body can be considered at its center, therefore we calculate the electric field on the surface of a sphere with the radius of the Earth
r = 6 , 37 106 m
E = k q / r²
q = E r² / k
q =
q = 4.5 10⁵ C
Now let's calculate the charge on the planet with E = 222 N / c and radius
r = 0.6 r_ Earth
r = 0.6 6.37 10⁶ = 3.822 10⁶ m
E = k q / r²
q = E r² / k
q =
q = 3.6 10⁵ C