Question

A positively charged disc is placed on a horizontal plane. A charged particle is released from a certain height on its axis. The particle just reaches the centre of the disc.

• A. Particle has negative charge on it
• B. Total potential energy (gravitational + electrostatic) of the particle first increases, then decreases
• C. Total potential energy (gravitational + electrostatic) of the particle first decreases, then increases
• D. Total potential energy (gravitational + electrostatic) of the particle continuously decreases

Answer 1

Answer: (C)

Total potential energy (gravitational + electrostatic) of the particle first decreases, then increases

Answer 2

The problem describes a charged particle released from a height on the axis of a positively charged disc. The particle "just reaches" the center of the disc, implying its initial and final kinetic energies are zero. We need to determine the charge of the particle and the behavior of its total potential energy.

1. Determine the charge of the particle:

Let the origin be at the center of the disc. The height of the particle is $z$.

The gravitational potential energy is $U_g(z) = mgz$.

The electrostatic potential energy is $U_e(z) = qV(z)$, where $q$ is the charge of the particle and $V(z)$ is the electrostatic potential due to the disc at height $z$.

For a positively charged disc, the electrostatic potential $V(z)$ on its axis is positive and decreases as $z$ increases (from a maximum at $z=0$ to zero at $z \to \infty$). Therefore, $V(0) > V(h)$ for any $h > 0$.

The particle is released from rest at $z=h$ and just reaches the center ($z=0$) with zero velocity. By conservation of mechanical energy:

Total Initial Energy = Total Final Energy

$K_i + U_{g,i} + U_{e,i} = K_f + U_{g,f} + U_{e,f}$

$0 + mgh + qV(h) = 0 + mg(0) + qV(0)$

$mgh + qV(h) = qV(0)$

$mgh = q(V(0) - V(h))$

Since $mgh$ is positive (mass $m>0$, gravity $g>0$, height $h>0$) and $(V(0) - V(h))$ is positive, it implies that $q$ must be positive.

Therefore, the particle has a positive charge. This rules out the first option.

2. Analyze the total potential energy:

The total potential energy of the particle is $U_{total}(z) = U_g(z) + U_e(z) = mgz + qV(z)$.

We have established that $q > 0$.

The particle moves from $z=h$ to $z=0$.

The derivative of the total potential energy with respect to height $z$ is:

$\frac{dU_{total}}{dz} = mg + q \frac{dV}{dz}$

We know that the electric field $E_z = -\frac{dV}{dz}$. For a positively charged disc, the electric field $E_z$ on its axis points away from the disc (upwards for $z>0$). So, $E_z$ is positive.

Therefore, $\frac{dV}{dz} = -E_z$.

Substituting this into the derivative equation:

$\frac{dU_{total}}{dz} = mg - qE_z$

Let's analyze the sign of $\frac{dU_{total}}{dz}$ at the start and end points of the motion:

• At $z=h$ (initial position): The particle is released from rest and moves downwards. This means the net force on it must be downwards. The gravitational force $mg$ is downwards, and the electrostatic force $qE_z$ (since $q>0$) is upwards (repulsive). So, $mg - qE_z(h) > 0$. This implies $\frac{dU_{total}}{dz}(h) > 0$.
• At $z=0$ (final position): The particle "just reaches" the center and stops. This means it has decelerated to zero velocity. For this to happen, the net force must be upwards at $z=0$ (or at least during the deceleration phase leading to $z=0$). So, $mg - qE_z(0) < 0$. This implies $\frac{dU_{total}}{dz}(0) < 0$.

Since $\frac{dU_{total}}{dz}$ changes from positive at $z=h$ to negative at $z=0$, and $E_z$ is a continuous function of $z$, there must be some intermediate height $z_0$ (where $0 < z_0 < h$) at which $\frac{dU_{total}}{dz} = 0$. This point $z_0$ corresponds to a minimum in the total potential energy.

Now, let's trace the behavior of $U_{total}(z)$ as the particle moves from $z=h$ down to $z=0$:

• From $z=h$ to $z=z_0$: Since $\frac{dU_{total}}{dz} > 0$, as $z$ decreases, $U_{total}(z)$ decreases.
• From $z=z_0$ to $z=0$: Since $\frac{dU_{total}}{dz} < 0$, as $z$ decreases, $U_{total}(z)$ increases.

Therefore, the total potential energy (gravitational + electrostatic) of the particle first decreases, then increases.