Elastic Collisions, Protons, and Electric Potential of Rings and Disks

Elastic Collision and Closest Approach of Two Protons

• Problem Description:

  • Two protons travel directly towards each other and collide head-on in a perfectly elastic manner.

  • Initial State ($t_0$): The protons are infinitely far apart, heading directly at one another.

  • Final State ($t_y$): After the interaction, they move away and end up infinitely far apart again.

  • Goal: Determine the final velocities of the two protons when far apart and calculate their minimum separation distance ($r_{min}$).

• System Parameters and Physical Constants:

  • Mass of Proton ($m_p$): $1.67 \times 10^{-27}\,\text{kg}$.

  • Charge of Proton ($q_p$): $1.609 \times 10^{-19}\,\text{C}$ (also noted as $1.602 \times 10^{-19}\,\text{C}$ in calculation steps).

  • Initial Velocity of Proton A ($v_{A1}$): $1.98 \times 10^4\,\text{m/s} \hat{i}$.

  • Initial Velocity of Proton B ($v_{B1}$): $-8.62 \times 10^4\,\text{m/s} \hat{i}$.

  • Coulomb Constant ($k$): $8.99 \times 10^9\,\text{N}\,\text{m}^2\,\text{C}^{-2}$.

• Conservation Principles:

  • Linear Momentum: The system consists of two protons with no external forces. Gravity is explicitly identified as being too weak to have a measurable effect.

  • Equation: $m_p v_{A1} + m_p v_{B1} = m_p v_{A4} + m_p v_{B4}$.

  • Simplified (since masses cancel): $v_{A1} + v_{B1} = v_{A4} + v_{B4}$.

  • Total Energy: Since it is an elastic collision, kinetic energy ($K$) and potential energy ($U$) are conserved.

  • Initial energy is purely kinetic ($U \approx 0$ at infinity).

• Part (a): Final Velocities ($v_{A4}, v_{B4}$):

  • Step 1: Momentum Conservation:

  • $1.98 \times 10^4 + (-8.62 \times 10^4) = v_{A4} + v_{B4}$

  • $-6.64 \times 10^4\,\text{m/s} = v_{A4} + v_{B4}$

  • Expression for $v_{A4}$: $v_{A4} = -6.64 \times 10^4 - v_{B4}$.

  • Step 2: Kinetic Energy Conservation:

  • $\frac{1}{2} m_p (v_{A1})^2 + \frac{1}{2} m_p (v_{B1})^2 = \frac{1}{2} m_p (v_{A4})^2 + \frac{1}{2} m_p (v_{B4})^2$

  • $(1.98 \times 10^4)^2 + (-8.62 \times 10^4)^2 = (v_{A4})^2 + (v_{B4})^2$

  • $3.92 \times 10^8 + 74.30 \times 10^8 = 7.82 \times 10^9\,\text{m}^2\,\text{s}^{-2} = (v_{A4})^2 + (v_{B4})^2$.

  • Step 3: Solving the Quadratic System:

  • Substitute $v_{A4}$ into the energy equation:

  • $7.82 \times 10^9 = (-6.64 \times 10^4 - v_{B4})^2 + (v_{B4})^2$

  • $7.82 \times 10^9 = 4.41 \times 10^9 + 1.328 \times 10^5 v_{B4} + v_{B4}^2 + v_{B4}^2$

  • $2 v_{B4}^2 + 1.328 \times 10^5 v_{B4} - 3.41 \times 10^9 = 0$.

  • Step 4: Results:

  • Solving the quadratic provides two solutions: the initial velocities and the final swapped velocities.

  • Final Velocity Proton A ($v_{A4}$): $-8.62 \times 10^4\,\text{m/s}$.

  • Final Velocity Proton B ($v_{B4}$): $1.98 \times 10^4\,\text{m/s}$.

• Part (b): Minimum Separation ($r_{min}$):

  • Condition for Closest Approach: This occurs when the relative velocity is zero. At time $t_2$, both protons travel at the same velocity ($v_{Az} = v_{Bz} = v_z$).

  • Calculating $v_z$:

  • Using momentum: $m_p (v_{A1} + v_{B1}) = (m_p + m_p) v_z$

  • $v_z = \frac{-6.64 \times 10^4}{2} = -3.32 \times 10^4\,\text{m/s}$.

  • Energy Balance at Closest Approach:

  • $K_{initial} = K_{t2} + U_{t2}$

  • $K_{initial} = 6.53 \times 10^{-18}\,\text{J}$ (Calculated as $3.27 \times 10^{-19} + 6.204 \times 10^{-18}$).

  • Kinetic Energy at $t_2$ ($K_{t2}$): $\frac{1}{2}(2 m_p)(v_z)^2 = (1.67 \times 10^{-27})(-3.32 \times 10^4)^2 = 1.84 \times 10^{-18}\,\text{J}$.

  • Potential Energy at $t_2$ ($U_{t2}$): $\frac{k q_p^2}{r_{min}} = \frac{(8.99 \times 10^9)(1.602 \times 10^{-19})^2}{r_{min}}$.

  • Solving for $r_{min}$:

  • $6.53 \times 10^{-18} = 1.84 \times 10^{-18} + \frac{2.31 \times 10^{-28}}{r_{min}}$

  • $4.69 \times 10^{-18} = \frac{2.31 \times 10^{-28}}{r_{min}}$

  • $r_{min} = 4.92 \times 10^{-11}\,\text{m}$.

Electric Potential of a Uniformly Charged Ring

• Problem Setup:

  • A ring of radius $R$ carries a total charge $Q$.

  • The ring is centered at the origin in the $x-y$ plane.

  • Field Point ($P$): Located at $(0, 0, z_0)$ on the $z$-axis.

• Symmetry Analysis:

  • The electric potential ($V$) is NOT expected to be zero.

  • For $V=0$ due to symmetry, the field point must be halfway between identical shapes with equal magnitude of charge but opposite signs.

• Derivation Process:

  • Identify Charge Element ($dq$): The charge is 1D, so break it into points along the ring.

  • $dq = \lambda ds = Q \frac{ds}{2\pi R}$.

  • In polar coordinates: $ds = R d\theta$, so $dq = \frac{Q}{2\pi} d\theta$.

  • Distance ($r$): The distance from any point $(x, y, 0)$ on the ring to point $P(0, 0, z_0)$.

  • vector $\mathbf{r} = (0-x)\mathbf{i} + (0-y)\mathbf{j} + (z_0-0)\mathbf{k}$.

  • magnitude $|r| = \sqrt{x^2 + y^2 + z_0^2}$.

  • Since $x^2 + y^2 = R^2$ on the ring, $r = \sqrt{R^2 + z_0^2}$.

  • Integration:

  • $dV = \frac{k dq}{r} = \frac{k (Q / 2\pi) d\theta}{\sqrt{R^2 + z_0^2}}$

  • $V = \int_0^{2\pi} \frac{k Q}{2\pi \sqrt{R^2 + z_0^2}} d\theta$

  • $V = \frac{k Q}{2\pi \sqrt{R^2 + z_0^2}} [\theta]_0^{2\pi} = \frac{k Q}{\sqrt{R^2 + z^2}}$.

• Verification and Limits:

  • Point Charge Limit: If $z \gg R$, the denominator $\sqrt{R^2 + z^2} \approx z$.

  • Result: $V \approx \frac{k Q}{z}$, which is the potential of a point charge at distance $z$.

  • Electric Field Check: The field $E_z$ can be derived from the negative derivative of the potential.

  • $E_z = -\frac{dV}{dz} = -kQ \frac{d}{dz} (R^2 + z^2)^{-1/2}$

  • $E_z = -kQ [-\frac{1}{2}(R^2 + z^2)^{-3/2}(2z)] = \frac{k Q z}{(R^2 + z^2)^{3/2}}$.

  • This matches the expected electric field formula derived in previous classes.

Electric Potential of a Uniformly Charged Disk

• Problem Setup:

  • A flat disk of radius $R$ carries a total charge $Q$ uniformly distributed over its area.

  • The disk is centered at the origin in the $x-y$ plane.

  • Goal: Find the potential at point $P(0, 0, z_0)$.

• Method: Summation of Rings:

  • A disk is treated as a collection of thin rings with radius $r$ and thickness $dr$.

  • Area Element ($dA$): $dA = 2\pi r dr$.

  • Charge Element ($dq$): Based on the fraction of the total area.

  • $dq = Q \left( \frac{dA}{A_{total}} \right) = Q \frac{2\pi r dr}{\pi R^2} = \frac{2Q r dr}{R^2}$.

• Integration:

  • Using the potential formula for a single ring from the previous section: $dV = \frac{k dq}{\sqrt{r^2 + z^2}}$.

  • $dV = \frac{k (2Q r dr / R^2)}{\sqrt{r^2 + z^2}}$.

  • Integrate from $r=0$ to $r=R$:

  • $V = \frac{2kQ}{R^2} \int_0^R \frac{r dr}{\sqrt{r^2 + z^2}}$.

  • Substitution: Let $u = r^2 + z^2$, then $du = 2r dr$.

  • $V = \frac{2kQ}{R^2} [\sqrt{r^2 + z^2}]_0^R$

  • Final Formula: $V = \frac{2kQ}{R^2} (\sqrt{R^{2} + z^{2}} - z)$.

• Verification against Electric Field:

  • The electric field $E_z$ is the negative derivative of the potential with respect to $z$:

  • $E = -\frac{dV}{dz} = -\frac{2kQ}{R^2} \frac{d}{dz} [ (R^2 + z^2)^{1/2} - z ]$

  • $E = -\frac{2kQ}{R^2} [ \frac{1}{2}(R^2 + z^2)^{-1/2}(2z) - 1 ]$

  • $E = \frac{2kQ}{R^2} [ 1 - \frac{z}{\sqrt{R^2 + z^2}} ]$.

  • This reproduces the established formula for the electric field of a charged disk.

Questions & Discussion

• Q: Why do we use $t_y$ and $t_z$?

  • A: These time steps represent specific phases of the interaction. $t_0$ is the start (infinite distance), $t_1$ is just before interaction, $t_2$ is the point of closest approach (velocities equalized), and $t_y$ is the reprise of the infinite distance state after the collision.

• Q: Is gravity included?

  • A: No, the transcript explicitly states gravity is too weak to have any effect in the context of these proton-proton interactions.

• Q: How do we handle symmetry in potential vs. field?

  • A: Potential is a scalar, so we do not need to calculate vector components. However, symmetry is still useful to determine if the potential might be zero. For the ring and disk, the potential remains non-zero on the $z$-axis because all charge elements are at the same distance and have the same sign.