A non-conducting disc of radius a and uniform positive surface charge density sigma is placed on the

Q: Why is the final kinetic energy zero?

The problem states the particle "just reaches the disc". This implies its velocity becomes zero exactly as it touches the disc, meaning all its initial gravitational potential energy has been converted into electrostatic potential energy.

Q: How do we get the potential of the disc on its axis?

It is derived by integrating the potential of elemental rings over the disc from radius $0$ to $a$. The result is a standard formula: $V = \frac{\sigma}{2\varepsilon_0}(\sqrt{a^2+H^2}-H)$.

Q: Why is the equilibrium stable?

At the equilibrium position $H = a/\sqrt{3}$, the total potential energy $U$ is at a local minimum. A particle displaced slightly from a minimum potential energy position experiences a restoring force, which is the condition for stable equilibrium.

Q: What does the given relation $q/m = 4\varepsilon_0 g/\sigma$ signify?

It is a specific condition provided to simplify the algebra. By substituting this relation, the electrostatic force terms and gravitational force terms become comparable, allowing us to solve for $H$ cleanly.

Q: Why does the potential energy graph go up after the minimum?

As the particle moves further away from the disc, the electrostatic repulsion weakens, but the gravitational potential energy ($mgH$) increases linearly. At large distances, the linear increase of $mgH$ dominates, causing the total potential energy to rise.

Question

A non-conducting disc of radius $a$ and uniform positive surface charge density $\sigma$ is placed on the ground with its axis vertical. A particle of mass $m$ and positive charge $q$ is dropped, along the axis of the disc from a height $H$ with zero initial velocity. The particle has $q/m = 4\varepsilon_0 g/\sigma$.
(a) Find the value of $H$ if the particle just reaches the disc.
(b) Sketch the potential energy of the particle as a function of its height and find its equilibrium position.

Sigma Percentile · Accepted Answer

### The Dance of Gravity and Electrostatics

Imagine a positively charged particle suspended high above a positively charged disc. Gravity relentlessly pulls the particle downward, while the electrostatic force from the disc pushes it upward. This problem is a beautiful exploration of how these two fundamental forces interact, balance, and dictate the motion of the particle.

### Analyzing the Setup

We are given a non-conducting disc of radius $a$ with a uniform positive surface charge density $\sigma$. A particle of mass $m$ and positive charge $q$ is dropped from a height $H$ along the central axis of the disc.

The problem provides a very specific and helpful relation:

$$\frac{q}{m} = \frac{4\varepsilon_0 g}{\sigma}$$

This relation is the key to simplifying the complex algebra that will follow. It essentially links the electrostatic parameters ($q, \sigma, \varepsilon_0$) with the gravitational parameters ($m, g$).

### The Master Equation: Conservation of Energy

For part (a), we are told the particle is dropped from rest and *just* reaches the disc. This means its initial velocity is zero, and its final velocity at the disc's center is also zero. Since there is no change in kinetic energy ($\Delta K = 0$), the mechanical energy is perfectly conserved between the gravitational and electrostatic potential energies.

The decrease in gravitational potential energy must exactly equal the increase in electrostatic potential energy:

$$mgH = q[V(0) - V(H)]$$

We know the electric potential $V$ on the axis of a uniformly charged disc at a distance $H$ is:

$$V(H) = \frac{\sigma}{2\varepsilon_0} \left[ \sqrt{a^2 + H^2} - H ight]$$

At the center of the disc ($H = 0$), this simplifies to:

$$V(0) = \frac{\sigma a}{2\varepsilon_0}$$

### Final Calculation for Height $H$

Substituting these potential expressions into our energy conservation equation, we get:

$$mgH = q \left[ \frac{\sigma a}{2\varepsilon_0} - \frac{\sigma}{2\varepsilon_0} (\sqrt{a^2 + H^2} - H) ight]$$

Dividing by $m$ and factoring out the constants:

$$gH = \left(\frac{q}{m}ight) \frac{\sigma}{2\varepsilon_0} \left[ a - \sqrt{a^2 + H^2} + H ight]$$

Now, we use the magic relation provided in the problem: $\frac{q}{m} = \frac{4\varepsilon_0 g}{\sigma}$. Substituting this into the coefficient gives:

$$\left(\frac{4\varepsilon_0 g}{\sigma}ight) \frac{\sigma}{2\varepsilon_0} = 2g$$

Our equation beautifully simplifies to:

$$gH = 2g \left[ a - \sqrt{a^2 + H^2} + H ight]$$

Canceling $g$ and rearranging terms:

$$H = 2a - 2\sqrt{a^2 + H^2} + 2H$$

$$2\sqrt{a^2 + H^2} = 2a + H$$

Squaring both sides to eliminate the square root:

$$4(a^2 + H^2) = 4a^2 + H^2 + 4aH$$

$$4a^2 + 4H^2 = 4a^2 + H^2 + 4aH$$

$$3H^2 = 4aH$$

Since $H 
eq 0$, we can divide by $H$ to find our final answer for part (a):

**$$H = \frac{4}{3}a$$**

### Finding the Equilibrium Position

For part (b), we need to analyze the total potential energy $U(H)$ of the particle as a function of its height. The total potential energy is the sum of the electrostatic and gravitational potential energies:

$$U(H) = U_e + U_g = qV(H) + mgH$$

$$U(H) = \frac{q\sigma}{2\varepsilon_0} \left[ \sqrt{a^2 + H^2} - H ight] + mgH$$

Using our magic relation again, we know that $\frac{q\sigma}{2\varepsilon_0} = 2mg$. Substituting this in:

$$U(H) = 2mg \left[ \sqrt{a^2 + H^2} - H ight] + mgH$$

$$U(H) = mg \left[ 2\sqrt{a^2 + H^2} - 2H + H ight]$$

$$U(H) = mg \left[ 2\sqrt{a^2 + H^2} - H ight]$$

To find the equilibrium position, we must find where the net force is zero. The net force is the negative gradient of the potential energy ($F = -\frac{dU}{dH}$). Setting the derivative to zero:

$$\frac{dU}{dH} = mg \left[ 2 \cdot \frac{1}{2\sqrt{a^2 + H^2}} \cdot 2H - 1 ight] = 0$$

$$\frac{2H}{\sqrt{a^2 + H^2}} - 1 = 0$$

$$2H = \sqrt{a^2 + H^2}$$

Squaring both sides:

$$4H^2 = a^2 + H^2$$

$$3H^2 = a^2$$

**$$H = \frac{a}{\sqrt{3}}$$**

At this height, the potential energy is at a local minimum, which means this is a position of **stable equilibrium**. If the particle is slightly displaced from this height, it will experience a restoring force pushing it back.

Visualized Solution

$Visualizing the Setup$

$Electric Potential on the Axis of a Disc$

$Conservation of Mechanical Energy$

$Solving for H$

$Algebraic Manipulation$

$Total Potential Energy Function$

$Finding Equilibrium Position$

$Solving for Equilibrium Height$

The Sigma Insight: Electric Potential and Potential Difference

The Dance of Gravity and Electrostatics

Analyzing the Setup

The Master Equation

Final Calculation for Height $H$

Finding the Equilibrium Position

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Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Visualizing the Setup

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Electric Potential on the Axis of a Disc

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conservation of Mechanical Energy

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for H

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Algebraic Manipulation

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Total Potential Energy Function

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Finding Equilibrium Position

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Solving for Equilibrium Height