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

$V\left(z\right)=\frac{\sigma }{2{\epsilon }_0}(\sqrt{R^2+z^2}-z)$

A particle of positive charge q is placed initially at rest at a point on the z axis with z = z₀ and z₀ > 0. In addition to the Coulomb force, the particle experiences a vertical force $\overrightarrow{F}=-c\hat{k}$ with c > 0. Let $\beta =\frac{2{\epsilon }_{0}c}{\sigma q}$. Which of the following statements (s) is (are) correct ?

The correct answers are: Option A (β = 1/4, z₀ = 25R/7, particle reaches origin), Option C (β = 1/4, z₀ = R/√3, particle returns to z₀), and Option D (β > 1, particle always reaches origin).

We will use energy methods. The total potential energy of the particle (charge q) at height z on the z-axis consists of two parts: the Coulomb potential energy and the potential energy due to the constant downward force F = −ck̂.

Step 1: Set Up the Total Potential Energy U(z)

The Coulomb potential is given as:

$V\left(z\right)=\frac{\sigma }{2{\epsilon }_0}(\sqrt{R^2+z^2}-z)$

The Coulomb potential energy of charge q is U_E = qV(z):

$U_E\left(z\right)=\frac{\sigma }{2{\epsilon }_0}q(\sqrt{R^2+z^2}-z)$

The force F = −ck̂ is a constant downward force. Its potential energy is:

$U_F\left(z\right)=cz$

(Since F = −dU_F/dz = −c, so U_F = cz.)

The total potential energy is:

$U\left(z\right)=\frac{\sigma }{2{\epsilon }_0}q(\sqrt{R^2+z^2}-z)+cz$

Let us factor out $\frac{\sigma }{2{\epsilon }_0}q$ and introduce $\beta =\frac{2{\epsilon }_{0}c}{\sigma q}$, so that $c=\frac{\beta }{\sigma }$:

$U\left(z\right)=\frac{\sigma q}{2{\epsilon }_0}[\sqrt{R^2+z^2}-z+\beta z]$

$U\left(z\right)=\frac{\sigma q}{2{\epsilon }_0}[\sqrt{R^2+z^2}+(\beta -1\left)z\right]$

Step 2: Energy Conservation and the Condition for Reaching the Origin

The particle starts from rest at z = z₀. By energy conservation (kinetic energy = 0 at start), total mechanical energy E = U(z₀).

The particle can reach z = 0 only if U(0) ≤ U(z₀) (so it doesn't run out of energy). At z = 0:

$U\left(0\right)=\frac{\sigma q}{2{\epsilon }_0}R$

At z = z₀:

$U\left(z_0\right)=\frac{\sigma q}{2{\epsilon }_0}[\sqrt{R^2+z_0^2}+(\beta -1\left)z_0\right]$

The condition for reaching the origin is U(z₀) ≥ U(0):

$\sqrt{R^2+z_0^2}+(\beta -1)z_0\ge R$

But the particle might be stopped by a potential barrier in between. So we need to analyze U(z) more carefully. Let's find the critical point(s) of U(z) for z > 0.

Step 3: Finding Critical Points of U(z)

Differentiate U(z) with respect to z and set to zero:

$\frac{dU}{dz}=\frac{\sigma q}{2{\epsilon }_0}[\frac{z}{\sqrt{R^2+z^2}}+(\beta -1\left)\right]=0$

This gives:

$\frac{z}{\sqrt{R^2+z^2}}=1-\beta$

This equation has a solution only when the right-hand side is positive (i.e., β < 1) and lies in [0,1). For β ≥ 1, the left side is always positive and the right side ≤ 0, so dU/dz > 0 always — meaning U(z) is monotonically increasing for z > 0, and thus the particle will always roll "down" the energy hill from z₀ toward z = 0. This proves Option D is correct.

For 0 < β < 1, squaring both sides:

$\frac{z^2}{R^2+z^2}={(1-\beta )}^2$

$z^2=(1-\beta {)}^2(R^2+z^2)$

$z^2[1-(1-\beta {)}^2]=(1-\beta {)}^2{R}^2$

$z^2\beta (2-\beta )=(1-\beta {)}^2{R}^2$

$z_c=\frac{(1-\beta )R}{\sqrt{\beta (2-\beta )}}$

This is the location of the potential barrier (local maximum of U for 0 < β < 1) at z = z_c.

Step 4: Compute U at the barrier z_c

At z = z_c, we have $\frac{z_c}{\sqrt{R^2+z_c^2}}=1-\beta$, so:

$\sqrt{R^2+z_c^2}=\frac{z_c}{1-\beta }$

Substituting into U:

$U\left(z_c\right)=\frac{\sigma q}{2{\epsilon }_0}[\frac{z_c}{1-\beta }+(\beta -1\left)z_c\right]$

$=\frac{\sigma q}{2{\epsilon }_0}{z}_c[\frac{1}{1-\beta }-(1-\beta \left)\right]$

$=\frac{\sigma q}{2{\epsilon }_0}{z}_c·\frac{1-(1-\beta {)}^2}{1-\beta }$

$=\frac{\sigma q}{2{\epsilon }_0}{z}_c·\frac{\beta (2-\beta )}{1-\beta }$

Now substituting z_c:

$U\left(z_c\right)=\frac{\sigma q}{2{\epsilon }_0}·\frac{(1-\beta )R}{\sqrt{\beta (2-\beta )}}·\frac{\beta (2-\beta )}{1-\beta }$

$U\left(z_c\right)=\frac{\sigma q}{2{\epsilon }_0}·R·\sqrt{\beta (2-\beta )}$

Let us define the natural energy scale $A=\frac{\sigma q}{2{\epsilon }_0}R$. Then:

$U\left(z_c\right)=A\sqrt{\beta (2-\beta )}$

And recall U(0) = A.

Step 5: Rule — when does the particle reach the origin?

The particle at z₀ (with U(z₀)) must pass over the barrier at z_c. It reaches the origin if and only if:

U(z₀) ≥ U(z_c) (enough energy to clear the barrier)

AND

U(z₀) ≥ U(0) (more energy than at origin, already guaranteed if first condition holds since U(z_c) ≥ U(0) for β < 1 — let's verify: is $\sqrt{\beta (2-\beta )}$ ≥ 1? Since β(2-β) = 1-(1-β)² ≤ 1 for 0 < β < 1, we get U(z_c) ≤ A = U(0)... wait, so the barrier is lower than U(0)?)

Let's reconsider the shape of U(z). At z → ∞: the √(R²+z²) ≈ z, so U ≈ A[z/R + (β-1)z/R] = A·β·z/R → ∞. At z = 0: U(0) = A. The derivative dU/dz = 0 gives z_c as a critical point. Is it a local minimum or maximum?

Since dU/dz = A/R · [z/√(R²+z²) + (β-1)] and at z = 0 this equals A(β-1)/R < 0 for β < 1, U is initially decreasing from U(0) = A. The critical point at z_c must then be a local minimum. U increases back to infinity as z → ∞.

So the shape of U(z) for 0 < β < 1 is: starts at A, decreases to a local minimum at z_c, then increases to ∞. This means the particle at z₀ > 0 starts in the "rising" region if z₀ > z_c, or in the "valley" if z₀ < z_c.

The particle starts at rest at z₀, so its total energy is E = U(z₀). It moves under the net force −dU/dz. It will move toward decreasing U. Since U decreases from z = 0 toward z_c and increases from z_c toward ∞:

• If z₀ > z_c: The particle initially moves toward z_c (toward smaller z, since U decreases in that direction). It will pass through z_c and then go toward z = 0 (U keeps decreasing from z_c to 0)... wait, U decreases from z = 0 to z = z_c, meaning U at z < z_c is greater than U at z_c.

Let me reclarify: dU/dz < 0 for 0 < z < z_c (U decreasing as z increases toward z_c), and dU/dz > 0 for z > z_c (U increasing). So U has a minimum at z_c, with U(0) = A > U(z_c) = A√(β(2-β)).

The force on the particle is F = −dU/dz:

• For 0 < z < z_c: dU/dz < 0 so F > 0 (particle pushed toward +z, i.e., away from origin)
• For z > z_c: dU/dz > 0 so F < 0 (particle pushed toward −z, i.e., toward z_c)

So z_c is a stable equilibrium. For the particle to reach the origin from z₀:

• If z₀ > z_c: The particle is pushed toward z_c (equilibrium). It will oscillate around z_c if it doesn't have enough energy to reach z = 0. It can only reach z = 0 if U(z₀) ≥ U(0) = A, since it must overcome the rising potential from z_c to 0.
• If z₀ < z_c: The particle is immediately pushed toward +z (away from origin). It will move to z_c, then oscillate. It cannot reach the origin unless... wait, if z₀ < z_c the particle is pushed away from the origin (F > 0), so it can never reach z = 0.
• If z₀ = z_c: Unstable case, particle at rest at local minimum → it stays there.

Summary of conditions to reach origin: The particle reaches z = 0 if and only if z₀ > z_c AND U(z₀) ≥ U(0) = A.

The condition U(z₀) ≥ A means:

$\sqrt{R^2+z_0^2}+(\beta -1)z_0\ge R$

$\sqrt{R^2+z_0^2}\ge R+(1-\beta )z_0$

Squaring (both sides positive for small enough z₀ or large enough β):

$R^2+z_0^2\ge R^2+2R(1-\beta )z_0+(1-\beta {)}^2{z}_0^2$

$z_0^2[1-(1-\beta {)}^2]\ge 2R(1-\beta )z_0$

$z_0^2·\beta (2-\beta )\ge 2R(1-\beta )z_0$

$z_0\ge \frac{2R(1-\beta )}{\beta (2-\beta )}$

Step 6: Evaluate Option A — β = 1/4, z₀ = 25R/7

Compute the threshold z₀ for reaching origin:

$z_0\ge \frac{2R(1-\frac{1}{4})}{\frac{1}{4}(2-\frac{1}{4})}=\frac{2R·\frac{3}{4}}{\frac{1}{4}}$

Let me compute step by step:

Numerator: 2R(1 - 1/4) = 2R · 3/4 = 3R/2

Denominator: (1/4)(2 - 1/4) = (1/4)(7/4) = 7/16

$z_0\ge \frac{3R/2}{7/16}=\frac{3R}{2}·\frac{16}{7}=\frac{24R}{7}$

Now check also that z₀ = 25R/7 > z_c:

$z_c=\frac{(1-\frac{1}{4})R}{\sqrt{\frac{1}{4}(2-\frac{1}{4})}}=\frac{\frac{3}{4}R}{\sqrt{\frac{7}{16}}}=\frac{\frac{3}{4}R}{\frac{\sqrt{7}}{4}}=\frac{3R}{\sqrt{7}}$

z_c = 3R/√7 ≈ 3R/2.646 ≈ 1.134R. And z₀ = 25R/7 ≈ 3.57R > z_c. ✓

Since z₀ = 25R/7 ≥ 24R/7 (the threshold), the particle does reach the origin. ✓ Option A is correct.

Step 7: Evaluate Option B — β = 1/4, z₀ = 3R/7

We need z₀ > z_c = 3R/√7 ≈ 1.134R, but z₀ = 3R/7 ≈ 0.429R < z_c.

Since z₀ < z_c, the net force on the particle at z₀ is directed away from the origin (toward +z). The particle moves away from the origin and cannot reach z = 0. Option B is incorrect.

Step 8: Evaluate Option C — β = 1/4, z₀ = R/√3

Check if z₀ = R/√3 ≈ 0.577R > z_c ≈ 1.134R. No! R/√3 < z_c.

Since z₀ < z_c, the particle is in the region where the net force points toward +z (away from origin). The particle will move away from z₀ in the +z direction, reach some maximum z, and return to z₀ (since U → ∞ as z → ∞ and by energy conservation it returns to same potential energy level, i.e., z₀). This means the particle returns back to z = z₀. ✓ Option C is correct.

Step 9: Evaluate Option D — β > 1, z₀ > 0

When β > 1, we showed that dU/dz = A/R · [z/√(R²+z²) + (β-1)] > 0 for all z > 0 (since both terms are positive). So U(z) is strictly increasing for all z > 0, meaning U(z₀) > U(0) for any z₀ > 0. Also, U is decreasing toward z = 0, so the particle always rolls downhill toward z = 0 and always reaches the origin. Option D is correct. ✓

Final Summary

The analysis rests on the shape of the total potential energy:

$U\left(z\right)=\frac{\sigma q}{2{\epsilon }_0}[\sqrt{R^2+z^2}+(\beta -1\left)z\right]$

• Option A ✓: β = 1/4, z₀ = 25R/7 > threshold 24R/7 and z₀ > z_c → particle reaches origin.
• Option B ✗: β = 1/4, z₀ = 3R/7 < z_c → particle is pushed away from origin, cannot reach it.
• Option C ✓: β = 1/4, z₀ = R/√3 < z_c → force pushes particle away from origin, it oscillates back to z₀.
• Option D ✓: β > 1 → U(z) monotonically decreasing toward z = 0 for all z > 0, so particle always reaches origin.