(II) A 12.0-cm-radius thin ring carries a uniformly distributed charge. A small 7.5-g sphere with a charge of is placed exactly at the center of the ring and given a very small push so it moves along the ring axis ( axis). How fast will the sphere be moving when it is 2.0 m from the center of the ring (ignore gravity)

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

a

explanation:

no diagram was provided, but i'm guessing it's the one i've attached to my answer.

here, a mechanical force turns the rod.   this generates an electric current in the coil, which powers the bulb.

the assestment will probley apele with cause and effect