Respuesta :
Answer:
W = 1.611 10⁻⁵ (1 / 0.250 - 1 / R_max)
If R_max is infinite, W = 6.444 10⁻⁵-5 J
Explanation:
The work is defined as
W = F.dr
In this case the outside is repulsive since the two bodies have the same charge signand the angle is zero
The force is given by Coulomb's law
F = k Q q / r²
The charge can be found from the density
ρ = Q / V
ρ = Q / (4/3 π R³)
Q = ρ 4/3 π R³
The work is
W = k ρ 4/3 π R³ q ∫ 1 / r² dr
We integrate
W = 8,988 10⁹ 7.40 10⁻⁹ 4/3 π 0.250³ 3.70 10⁻⁶ (-1 / r)
We evaluate between the lower limit r = R and the upper limit r = R_max
W = 16.11 10⁻⁶ (-1 / R_max + 1 / R)
W = 1.611 10⁻⁵ (1 / 0.250 - 1 / R_max)
If R_max is infinite, the work is
W = 1.611 10⁻⁵ 1 / 0.250
W = 6.444 10⁻⁵-5 J
Work (W) is = 1.611 10⁻⁵ (1 / 0.250 - 1 / R_max)
When R_max is infinite, W is = 6.444 10⁻⁵-5 J
How to define Uniform charge density?
Now The work is defined as
W is = F.dr
In this Situation the outside is repulsive hence the two bodies have the same charge sign and the angle is zero
Then The force is given by Coulomb's law is
F is = k Q q / r²
Then The charge can be found from the density
ρ is = Q / V
ρ is = Q / (4/3 π R³)
Q is = ρ 4/3 π R³
After that The work is
W is = k ρ 4/3 π R³ q ∫ 1 / r² dr
Then We integrate
W is = 8,988 10⁹ 7.40 10⁻⁹ 4/3 π 0.250³ 3.70 10⁻⁶ (-1 / r)
After that We evaluate between the lower limit r is = R and the upper limit r is = R_max
Then W is = 16.11 10⁻⁶ (-1 / R_max + 1 / R)
Now W is = 1.611 10⁻⁵ (1 / 0.250 - 1 / R_max)
When R_max is infinite, the work is
W is = 1.611 10⁻⁵ 1 / 0.250
W is = 6.444 10⁻⁵-5 J
Find more information about Uniform charge density here:
https://brainly.com/question/16687905
