Answer:
The kinetic energy K of the moving charge is K = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd
Explanation:
The potential energy due to two charges q₁ and q₂ at a distance d from each other is given by U = kq₁q₂/r.
Now, for the two charges q₁ = q₂ = Q separated by a distance d, the initial potential energy is U₁ = kQ²/d. The initial kinetic energy of the system K₁ = 0 since there is no motion of the charges initially. When the moving charge is at a distance of r = 3d, the potential energy of the system is U₂ = kQ²/3d and the kinetic energy is K₂.
From the law of conservation of energy, U₁ + K₁ = U₂ + K₂
So, kQ²/d + 0 = kQ²/3d + K
K₂ = kQ²/d - kQ²/3d = 2kQ²/3d
So, the kinetic energy K₂ of the moving charge is K₂ = 2kQ²/3d = 2Q²/(4πε)3d = Q²/6πεd
Answer:
v = 2 v₁ v₂ / (v₁ + v₂)
Explanation:
The body travels the first half of the distance with velocity v₁. The time it takes is:
t₁ = (d/2) / v₁
t₁ = d / (2v₁)
Similarly, the body travels the second half with velocity v₂, so the time is:
t₂ = (d/2) / v₂
t₂ = d / (2v₂)
The average velocity is the total displacement over total time:
v = d / t
v = d / (t₁ + t₂)
v = d / (d / (2v₁) + d / (2v₂))
v = d / (d/2 (1/v₁ + 1/v₂))
v = 2 / (1/v₁ + 1/v₂)
v = 2 / ((v₁ + v₂) / (v₁ v₂))
v = 2 v₁ v₂ / (v₁ + v₂)
Answer:
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Answer:
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