Electric Physics

Question

A proton is held fixed at the origin. A second proton is placed on the x axis at x = 1.5 m and then released. At what point will this proton's potential energy be equal to its kinetic energy? Express your answer with the appropriate units.

Answer 1

We are tasked with finding the specific point at which the potential energy of a proton is equal to its kinetic energy as it moves away from a fixed proton placed at the origin. Initially, the proton is at x =1.5m and then released. We need to determine where the Potential Energy (U) equals the Kinetic Energy (KE). We know that the potential energy of two protons r meters apart can be represented as

U = k*q₁q₂ /r

1. k is Coulomb's constant (~8.99×10⁹ Nm²C^-2)
2. q_1,2 the charge of a proton (it equals ~1.60×10^-19C for each)
3. r is the distance of separation (initially, this distance is 1.5m, but we need to find the distance where KE=U)

Since we know that energy is conserved (the total energy remains constant unless acted on by an external force), we have

U_initial = k*q₁q₂/(1.5meters) = U + KE

The problem asks us to find the distance where the kinetic energy equals the potential energy, or KE = U. So, we substitute KE = U into the equation:

k*q₁q₂/(1.5meters) = U + U = 2U

Now, using the formula for potential energy:

k*q₁q₂/(1.5meters) = 2 (k*q₁q₂ /r)

At this point, we can cancel out the constants k * q1 * q2 from both sides, which gives us:

1/(1.5meters) = 2/r

Solving for r, we get:

r = 2 * 1.5 = 3 meters

Thus, the protons will be 3 meters apart when the moving proton’s kinetic energy equals its potential energy.

Answer 2

Two positive protons repel each other, so the proton at distance of 1.5m will fly off in the positive x direction. The work done on the free proton is positive since the force and displacement are in the same direction. The work done on the proton will result in a loss of potential energy and an equivalent gain in kinetic energy. The work equation can be obtained with calculus and is found to be

W = kQ^2 [ -1 / R₂ + 1/R₁] where R1 is the initial separation of the charges, R2 the final separation, k the coulomb constant (k = 8.99 x 10^9 N m^2/C^2), and Q the charge on the proton (1.602 x 10^-19 C). In this case, the free proton is assumed to go to infinity, so 1/R2 = 0. The kinetic energy will then be given by

KE = kQ^2 [ 0 + 1/R₁] = 1.54 x 10^-28 Joules.