A 13.2 cm -radius thin ring carries a uniformly distributed 16.5 μC charge. A small 7.2 g sphere with a charge of 2.5 μC is placed exactly at the center of the ring and given a very small push so it moves along the ring axis (+x axis).
1 Answer | Add Yours
The potential of the electric field created by a charged ring is given by
`V = (kQ)/sqrt(r^2 + x^2)` , where x is the distance from the center of ring on the axis of the ring, Q is the charge and r is the radius of the ring, and `k = 1/(4piepsilon_0)` (`epsilon_0` is the dielectric constant.) (Please see the reference link to find out how this formula can be obtained.)
Since both ring and the sphere are positively charged, once the sphere is slightly pushed away from the center of the ring (where the net force on it was 0), it will experience the electric force that will push it further away from the ring. Once it starts moving, its potential energy is going to decrease. Then, its kinetic energy will increase and it will pick up speed.
The law of conservation of energy states
`DeltaK + DeltaU = 0`
The change of kinetic energy is `DeltaK = (mv^2)/2` , where v is the sphere's speed 2 meters away from the center of the ring. The initial kinetic energy is assumed to be 0 because the sphere is only given a slight push, but not a significant initial velocity.
The change of the potential energy is
`DeltaU = q(V(2) - V(0))` , where q is the charge of the sphere,
V(0) is the potential in the center of the ring: `V(0) = (kQ)/r` , and
V(2) is the potential 2 meters away: `V(2) = (kQ)/sqrt(r^2 + 4)`
Plugging everything in, we get
`(mv^2)/2 = -qkQ(1/sqrt(r^2 + 4) - 1/r)`
`(mv^2)/2 = 2.72 J`
`v = sqrt((2*2.72)/0.0072) = 27.5 m/s`
When 2 meters away from the center of the ring, the sphere will be moving with the speed of approximately 28 m/s.
We’ve answered 319,642 questions. We can answer yours, too.Ask a question