Answer:
v = 29.q m/s
Explanation:
From the question, the charge is uniformly distributed and therefore, each point in the x-axis is equidistant to every point of the ring.
we can compute the potential at a point (x, 0, 0) without resorting to integration:
15 µC/[4πεo√(x² + R²)]
The potential at the center of the ring is: (15 µC)/{(4πεo) x (12.0 cm)} = 1.123 x 10^(6) V.
However, 2m from the center of the ring, the potential gives:
(15 µC)/{(4πεo x√((2m)² + (12.0 cm)²} = 6.73 x 10^(4) V.
The potential difference is: 6.73 x 10^(4) - 1.123 x 10^(6) V = -1.06 x 10^(6) V.
The change in potential ΔU = qΔV = (3 µC)(-1.06 x 10^(6) V) = -3.17 J.
In conservation of energy, the kinetic energy change ΔK = -ΔU = 3.17 J. Since the particle starts from rest, the final KE = ΔK:
KE = 3.17J = (1/2)mv²
(Multiply both sides by2/m
3.17 x (2/m) = v²
(2 x 3.17 J)/(7.5 g) = v²
v² = 840 m²/s²
Take square root of both sides to get;
v = 29.1 m/s
