A uniform thin ring of charge, with radius 5.20 cm and total charge 6.60 ìC, is located in the yz-plane and centered on the origin. (In other words, the x-axis passes through the center of the ring, and is perpendicular to the ring.) A particle of mass 17.9 g and charge 4.80 ìC is on the negative x-axis, initially very far from the ring (effectively at .) The particle is launched in the positive x direction with some initial velocity, toward the center of the ring. Calculate the minimum initial velocity required for the particle to get past the center of the ring and come out the other side
To calculate the minimum initial velocity required for the particle to get past the center of the ring and come out the other side, we can use the concepts of energy conservation and the electric potential.
1. Start by determining the electric potential energy at the initial position. The electric potential energy (U) is given by:
U = k * (q1 * q2) / r
where k is the Coulomb's constant (k = 9.0 x 10^9 N m^2/C^2), q1 and q2 are the charges, and r is the distance between them.
In this case, q1 = 4.80 μC and q2 is the total charge of the ring (6.60 μC). The distance between them at the initial position is effectively infinite. Therefore, the electric potential energy can be approximated as zero.
2. At the initial position, the particle has both potential energy and kinetic energy. Therefore, the total mechanical energy (E) is given by:
E = U + K
where K is the kinetic energy of the particle.
3. As the particle moves towards the ring, the electric potential energy increases and the kinetic energy decreases. At the moment the particle is past the center of the ring, the electric potential energy is maximum and the kinetic energy is minimum.
4. At the center of the ring, the entire potential energy is converted into kinetic energy. The maximum potential energy is given by:
U_max = k * (q1 * q2) / (2 * r)
where r is the radius of the ring.
5. The minimum initial kinetic energy required for the particle to get past the center of the ring is half of the maximum potential energy (U_max):
K_min = U_max / 2
6. The kinetic energy is related to the velocity (v) of the particle as:
K = (1/2) * m * v^2
where m is the mass of the particle.
7. Substitute the formula for the kinetic energy into the equation from step 5 and solve for the minimum initial velocity (v):
(1/2) * m * v^2 = U_max / 2
Simplifying the equation gives:
v^2 = (U_max / m)
v = sqrt(U_max / m)
8. Substitute the values for U_max and m into the equation and calculate the minimum initial velocity required for the particle to get past the center of the ring and come out the other side.
It is important to note that the signs of the charges and direction of motion play a role in the calculations. Be sure to use the correct signs and directions when substituting values into the equations.