A disk of radius R with uniform positive charge density sigma is placed on the xy plane with its cen

Q: Why can we use potential energy for the external force?

Because the external force is constant ($-c\hat{k}$), it is conservative! Just like gravity near Earth's surface, a constant force has a well-defined potential energy, which is simply $cz$ in this case.

Q: What happens if the particle doesn't have enough energy to reach the origin?

It will momentarily come to rest at some point $z > 0$ where its kinetic energy becomes zero, and then the net force will push it back towards its starting position.

Q: Why did we check $U_{eff}(z_0) \ge U_{eff}(0)$ instead of just integrating the force?

Integrating the net force from $z_0$ to $0$ gives the work done, which is exactly equal to the negative change in potential energy! Using the potential energy function is just a faster, more elegant way to do the same math.

Q: In Option D, why does $\beta > 1$ guarantee reaching the origin?

If $\beta > 1$, the constant downward force is so strong that it completely overpowers the maximum possible upward Coulomb repulsion at every single point. The net force is always directed towards the origin!

Q: Could the particle cross the origin and go to negative $z$?

Yes! If it reaches the origin with non-zero kinetic energy, it will cross it. However, the question only asks if it *reaches* the origin, so we only need to ensure its kinetic energy doesn't drop to zero before $z=0$.

Question

A disk of radius $R$ with uniform positive charge density $\sigma$ is placed on the $xy$ plane with its center at the origin. The Coulomb potential along the $z$-axis is
$V(z) = \frac{\sigma}{2\epsilon_0} \left( \sqrt{R^2 + z^2} - z \right)$
A particle of positive charge $q$ is placed initially at rest at a point on the $z$ axis with $z = z_0$ and $z_0 > 0$. In addition to the Coulomb force, the particle experiences a vertical force $\vec{F} = -c\hat{k}$ with $c > 0$. Let $\beta = \frac{2c\epsilon_0}{q\sigma}$. Which of the following statement(s) is(are) correct?

Options:

(A) For $\beta = \frac{1}{4}$ and $z_0 = \frac{25}{7}R$, the particle reaches the origin.

(B) For $\beta = \frac{1}{4}$ and $z_0 = \frac{3}{7}R$, the particle reaches the origin.

(C) For $\beta = \frac{1}{4}$ and $z_0 = \frac{R}{\sqrt{3}}$, the particle returns back to $z = z_0$.

(D) For $\beta > 1$ and $z_0 > 0$, the particle always reaches the origin.

Sigma Percentile · Accepted Answer

### Analyzing the Setup Imagine a uniformly charged disk lying flat on the $xy$-plane. A positively charged particle is placed on the $z$-axis at a height $z_0$. This particle is caught in a tug-of-war between two forces: 1. The **Coulomb Force** ($F_e$): The positively charged disk repels the positive particle upwards. 2. The **External Force** ($F_{ext}$): A constant force pulling the particle downwards towards the origin. Because both of these forces are conservative, we don't need to mess around with complex kinematic equations or integrations of acceleration. Instead, we can use the elegant principle of **Conservation of Energy**. ### The Master Equation: Effective Potential Energy To use energy conservation, we define an **effective potential energy** $U_{eff}(z)$ that accounts for both forces. The potential energy due to the Coulomb force is simply $qV(z)$. The potential energy due to the constant downward force $-c\hat{k}$ is $cz$ (since $F = -\frac{dU}{dz}$). $$U_{eff}(z) = qV(z) + cz$$ Substituting the given expression for $V(z)$: $$U_{eff}(z) = \frac{q\sigma}{2\epsilon_0} \left( \sqrt{R^2 + z^2} - z \right) + cz$$ The problem introduces a dimensionless constant $\beta = \frac{2c\epsilon_0}{q\sigma}$. Let's express $c$ in terms of $\beta$: $$c = \frac{q\sigma\beta}{2\epsilon_0}$$ Plugging this back into our effective potential energy equation, we get our master equation: $$U_{eff}(z) = \frac{q\sigma}{2\epsilon_0} \left( \sqrt{R^2 + z^2} - z + \beta z \right)$$ For the particle to reach the origin ($z=0$) starting from rest at $z_0$, its initial energy must be greater than or equal to the energy at the origin. In other words, we need $\Delta U = U_{eff}(z_0) - U_{eff}(0) \ge 0$. ### Evaluating the Options: A Game of Energy Let's test the given options one by one. **Checking Option A:** We are given $\beta = \frac{1}{4}$ and $z_0 = \frac{25}{7}R$. Let's calculate the initial and final potential energies. $$U_{eff}(z_0) = \frac{q\sigma}{2\epsilon_0} R \left( \sqrt{1 + \frac{625}{49}} - \frac{3}{4} \cdot \frac{25}{7} \right) = \frac{q\sigma}{2\epsilon_0} R \left( \frac{\sqrt{674}}{7} - \frac{75}{28} \right)$$ $$U_{eff}(0) = \frac{q\sigma}{2\epsilon_0} R$$ Subtracting the two gives us the energy difference: $$\Delta U = \frac{q\sigma}{2\epsilon_0} R \left( \frac{4\sqrt{674} - 103}{28} \right)$$ Since $4\sqrt{674} = \sqrt{10784}$ and $103 = \sqrt{10609}$, we can clearly see that $4\sqrt{674} > 103$. Therefore, $\Delta U > 0$. The particle has more than enough energy to reach the origin! Option A is correct. **Checking Option B:** Now, let's test $\beta = \frac{1}{4}$ and $z_0 = \frac{3}{7}R$. $$U_{eff}(z_0) = \frac{q\sigma}{2\epsilon_0} R \left( \sqrt{1 + \frac{9}{49}} - \frac{3}{4} \cdot \frac{3}{7} \right) = \frac{q\sigma}{2\epsilon_0} R \left( \frac{\sqrt{58}}{7} - \frac{9}{28} \right)$$ $$\Delta U = \frac{q\sigma}{2\epsilon_0} R \left( \frac{4\sqrt{58} - 37}{28} \right)$$ Here, $4\sqrt{58} = \sqrt{928}$ and $37 = \sqrt{1369}$. Since $\sqrt{928} < \sqrt{1369}$, $\Delta U < 0$. The particle will run out of kinetic energy before reaching the origin. Option B is incorrect. **Checking Option C:** For $\beta = \frac{1}{4}$ and $z_0 = \frac{R}{\sqrt{3}}$: $$U_{eff}(z_0) = \frac{q\sigma}{2\epsilon_0} R \left( \sqrt{1 + \frac{1}{3}} - \frac{3}{4\sqrt{3}} \right) = \frac{q\sigma}{2\epsilon_0} R \left( \frac{5}{4\sqrt{3}} \right)$$ $$\Delta U = \frac{q\sigma}{2\epsilon_0} R \left( \frac{5}{4\sqrt{3}} - 1 \right)$$ Since $5 < 4\sqrt{3} \approx 6.928$, $\Delta U < 0$. The particle stops before reaching the origin. Because the forces are purely conservative, it will oscillate and return back to its starting position $z_0$. Option C is correct. **Checking Option D:** What if $\beta > 1$? Let's look at the net force acting on the particle: $$F_{net} = -\frac{dU_{eff}}{dz} = -\frac{q\sigma}{2\epsilon_0} \left( \frac{z}{\sqrt{R^2 + z^2}} - 1 + \beta \right)$$ Notice the term inside the parenthesis. Since $\frac{z}{\sqrt{R^2 + z^2}}$ is always positive for $z > 0$, and $\beta > 1$ means $(\beta - 1) > 0$, the entire term is strictly positive. This means $F_{net}$ is always negative! The downward external force completely overpowers the upward Coulomb repulsion at every single point. The particle will inevitably be crushed into the origin. Option D is correct. ### The Final Verdict By systematically applying the principle of energy conservation and analyzing the effective potential, we have successfully decoded the particle's fate. The correct options are **A, C, and D**.

Select Answer:

Visualized Solution

Visualizing the Setup

Effective Potential Energy

Simplifying U_{eff}(z)

Checking Option A

Checking Option B

Checking Option C

Checking Option D

Conclusion

The Sigma Insight: Electric Potential and Potential Difference

Analyzing the Setup

The Master Equation

Evaluating the Options

The Final Verdict By systematically applying the principle of energy conservation and analyzing the effective potential, we have successfully decoded the particle's fate

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