Answer:
A. 79.3 keV
Explanation:
Because the procedure involves many steps for its resolution and it works faster on paper and pencil, the detailed solution of this exercise is attached as a scanned image of the procedure for review.
In the procedure, the initial values of the problem and the replacement of these values with the correct formulas for this process are taken into account.
Answer:
0.86 m
Explanation:
q₁ = magnitude of positive charge = 5 x 10⁻⁶ C
q₂ = magnitude of negative charge = 3 x 10⁻⁶ C
r = distance between the two charges = 0.250 m
d = distance of the location of third charge from negative charge
q = magnitude of charge on third charge
Using equilibrium of electric force on third charge
d = 0.86 m
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