By what factor will the Electrostatic Force between two charged objects change when the amount of charge on both objects doubles AND the distance between the two charged objects triples?
1 answer:
Answer:
F' = (4/9)F
Explanation:
The electrostatic force between two charged objects is given by Coulomb's Law:
F = kq₁q₂/r² -------------------- equation (1)
where,
F = Electrostatic Force
k = Coulomb's Constant
q₁ = magnitude of first charge
q₂ = magnitude of second charge
r = distance between charges
Now, when the charges and distance altered as follows:
q₁' = 2q₁
q₂' = 2q₂
r' = 3r
Then,
F' = kq₁'q₂'/r'²
F' = k(2q₁)(2q₂)/(3r)²
F' = (4/9)kq₁q₂/r²
using equation (1):
<u>F' = (4/9)F </u>
You might be interested in
<span>We are using the formula: v = rω
Please consider that they have the same tangential velocity at their edges.
Therefore we can equate: r1 ω1 = r2 ω2
And re-arrange: ω1 / ω2 = r2 / r1</span>
Option B Polar substances are not likely to dissolve to a significant degree in nonpolar solvents. For example, ionic compounds are insoluble in hexane.
i don't understand sorry ;(
Answer:
The current would stop
Explanation:
Electric currents are interesting because they carry little to no momentum. As soon as you remove a power source, the whole current halts.
The wall exerts a force of equal magnitude but in the opposite direction. So the force by the wall is 10 N to the right.