A solid sphere of radius R carries a charge Q+q distributed uniformly over its volume. A very small

Q: Why do we consider the distance from the center of the sphere for electrostatic potential energy?

For a uniformly charged solid sphere, the electric potential outside the sphere is the same as if all its charge were concentrated at its center. Since the piece falls from the surface outwards, we use the center as our reference point for distance.

Q: Why is the initial kinetic energy zero?

The problem states that the piece 'gets detached' and 'falls down'. This implies it starts from rest, so its initial velocity and kinetic energy are zero.

Q: Does the charge of the detached piece affect the remaining sphere's field?

The problem asks us to 'Assume the remaining portion to be spherical.' This means we can treat the remaining sphere as having charge $Q$ and still behaving like a perfect point charge from the perspective of the falling piece.

Q: Why do we add the losses in gravitational and electrostatic potential energies?

Both gravity and electrostatic repulsion are pushing the piece downwards. As it moves down, it loses both types of potential energy, and both contribute to increasing its kinetic energy.

Q: What if the piece had a negative charge?

If the piece had a negative charge, the electrostatic force would be attractive (upwards). It would gain electrostatic potential energy as it falls, which would decrease its final kinetic energy compared to a purely gravitational fall.

Question

A solid sphere of radius $R$ carries a charge $Q+q$ distributed uniformly over its volume. A very small point-like piece of it of mass $m$ gets detached from the bottom of the sphere and falls down vertically under gravity. This piece carries charge $q$. If it acquires a speed $v$ when it has fallen through a vertical height $y$ (see figure), then (Assume the remaining portion to be spherical.)

Options:

(A) $v^2 = y \left[ \frac{qQ}{4\pi\epsilon_0 R^2 ym} + g \right]$

(B) $v^2 = 2y \left[ \frac{qQ}{4\pi\epsilon_0 R(R+y)m} + g \right]$

(C) $v^2 = y \left[ \frac{qQ}{4\pi\epsilon_0 R(R+y)m} + g \right]$

(D) $v^2 = 2y \left[ \frac{qQR}{4\pi\epsilon_0 (R+y)^3 m} + g \right]$

Sigma Percentile · Accepted Answer

### Analyzing the Setup

Imagine a large solid sphere carrying a total charge of $Q+q$. Right at the bottom of this sphere, a tiny piece of mass $m$ and charge $q$ breaks off. This is a crucial detail: because the piece takes away a charge $q$, the remaining sphere is left with a net charge of exactly $Q$.

The problem asks us to assume that the remaining portion is still perfectly spherical. This allows us to treat the remaining sphere as a point charge $Q$ located at its center when calculating the electric field and potential outside of it.

The detached piece falls vertically downwards under the influence of gravity. As it falls through a distance $y$, it accelerates and acquires a velocity $v$. Our goal is to find the mathematical expression for this velocity.

### The Master Equation

Since the piece is falling under the influence of gravity and electrostatic repulsion—both of which are conservative forces—we can invoke the powerful principle of **Conservation of Mechanical Energy**.

This principle states that the total mechanical energy of the system remains constant. In other words, any gain in kinetic energy must come at the expense of potential energy.

$$ \Delta K + \Delta U = 0 $$

Or, more intuitively:
**Gain in Kinetic Energy = Loss in Gravitational Potential Energy + Loss in Electrostatic Potential Energy**

Let's break down these energy terms. The piece starts from rest, so its initial kinetic energy is zero. After falling a distance $y$, its kinetic energy is $\frac{1}{2}mv^2$. Therefore, the gain in kinetic energy is simply $\frac{1}{2}mv^2$.

The loss in gravitational potential energy is straightforward: as it falls a height $y$, it loses $mgy$ of gravitational potential energy.

### Calculating Electrostatic Potential Energy

Now, let's look at the electrostatic potential energy. Initially, the piece is at the bottom surface of the sphere, which is at a distance $R$ from the center. The initial electrostatic potential energy is:

$$ U_i = \frac{kQq}{R} $$

After falling a distance $y$, the piece is now at a total distance of $R+y$ from the center of the sphere. The final electrostatic potential energy is:

$$ U_f = \frac{kQq}{R+y} $$

The loss in electrostatic potential energy is the difference between the initial and final states:

$$ 	ext{Loss in } U_{	ext{elec}} = U_i - U_f = kQq \left[ \frac{1}{R} - \frac{1}{R+y} ight] $$

### Final Calculation

Now we plug everything back into our master energy equation:

$$ \frac{1}{2}mv^2 = mgy + kQq \left[ \frac{1}{R} - \frac{1}{R+y} ight] $$

Let's simplify the term inside the bracket by taking the common denominator:

$$ \frac{1}{R} - \frac{1}{R+y} = \frac{(R+y) - R}{R(R+y)} = \frac{y}{R(R+y)} $$

Substituting this back into our equation gives:

$$ \frac{1}{2}mv^2 = mgy + \frac{kQq y}{R(R+y)} $$

To isolate $v^2$, we multiply the entire equation by 2 and divide by $m$:

$$ v^2 = 2gy + \frac{2kQq y}{m R(R+y)} $$

We can factor out $2y$ from both terms on the right side to make it look cleaner:

$$ v^2 = 2y \left[ g + \frac{kQq}{m R(R+y)} ight] $$

Finally, we substitute the value of Coulomb's constant, $k = \frac{1}{4\pi\epsilon_0}$:

$$ v^2 = 2y \left[ \frac{qQ}{4\pi\epsilon_0 R(R+y)m} + g ight] $$

And there we have it! This perfectly matches option (b). By carefully tracking the energy transformations, we've elegantly solved for the velocity of the falling charged particle.

Select Answer:

Visualized Solution

Initial Setup

The Fall

Conservation of Energy

Initial and Final Energies

Energy Equation

Substituting Potential Energy

Simplifying the Expression

Isolating Velocity

Final Answer

The Sigma Insight: Electrostatic Potential Energy

Analyzing the Setup

The Master Equation

Calculating Electrostatic Potential Energy

Final Calculation

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