Q.A positively charged particle is released from rest in an uniform electric field. The electric potential energy of the charge

The key idea is that a charge in a uniform electric field experiences a constant force, and its electric potential energy changes linearly with displacement along the field direction.

1. For a uniform field $\vec{E}$, the force on a charge $q$ is $\vec{F} = q\vec{E}$, which is constant. The electric potential energy is $U = qV$, where $V$ is the electric potential.
2. As the positive charge moves in the direction of $\vec{E}$, it moves from a point of higher potential to lower potential. Since $q > 0$, the product $qV$ decreases.
3. The loss in potential energy equals the gain in kinetic energy. The potential energy $U$ decreases linearly with distance $d$ moved along the field: $\Delta U = -qEd$.

Why This Happens: The Concept of Electric Potential Energy

Electric potential energy is the energy a charge has due to its position in an electric field. For a uniform field $\vec{E}$, the potential energy of a charge $q$ at a point is $U = qV$, where $V$ is the electric potential at that point. The key relationship is that the electric field points in the direction of decreasing potential — from higher to lower potential.

For a positive charge, $U = qV$ means that if $V$ decreases, $U$ also decreases. So when a positive charge is released from rest, it naturally moves from higher potential to lower potential, and its potential energy drops.

Think of it like a ball on a hill: the ball (positive charge) rolls downhill (toward lower potential), losing gravitational potential energy. Here, the "hill" is the electric potential landscape, and the charge slides down it.

Step-by-Step Reasoning

1. Set up the situation.

   A uniform electric field $\vec{E}$ points in some fixed direction. A positive charge $q > 0$ is placed at rest at a point where the electric potential is $V_1$.

2. Write the initial potential energy.

   The electric potential energy at the start is $U_i = q V_1$.

3. What happens when released?

   The electric force on the charge is $\vec{F} = q \vec{E}$, which points in the same direction as $\vec{E}$ (since $q > 0$). So the charge accelerates along $\vec{E}$.

4. How does potential change along the motion?

   In a uniform field, the potential decreases linearly in the direction of $\vec{E}$. If the charge moves a distance $d$ along $\vec{E}$, the potential drops by $\Delta V = -E d$. So at the new position, $V_2 = V_1 - E d$, which is less than $V_1$.

5. Find the final potential energy.

   At the new position, $U_f = q V_2 = q (V_1 - E d) = q V_1 - q E d$.

6. Compare initial and final.

   Since $q E d > 0$, we have $U_f < U_i$. The potential energy has decreased by exactly $q E d$.

7. Where did the energy go?

   The lost potential energy converts entirely into kinetic energy. The charge speeds up, so its kinetic energy increases by the same amount $q E d$. Total mechanical energy (potential + kinetic) is conserved.

Method: Reasoning About Potential-Energy Change for a Charge Released in a Field

This method predicts whether a charge's electric potential energy rises or falls once it is released in a field, without needing any numbers.

Steps

Step 1: Identify the force and its direction

For a positive charge, the force is along $\vec{E}$; for a negative charge, it is opposite to $\vec{E}$. A charge released from rest always accelerates in the direction of this force.

Step 2: Recall that the field points from high to low potential

By definition, $\vec{E}$ points in the direction of decreasing potential $V$. So "moving along $\vec{E}$" always means "moving toward lower $V$", regardless of the sign of the charge.

Step 3: Combine with $U = qV$ to get the sign of $\Delta U$

If the charge moves toward lower $V$: for $q>0$, $U=qV$ decreases; for $q<0$, $U=qV$ increases, since multiplying a decreasing $V$ by a negative $q$ flips the sign of the change.

Step 4: Cross-check with energy conservation

A charge released from rest gains kinetic energy as it accelerates. Since total mechanical energy is conserved, the kinetic energy gained must equal the potential energy lost — confirming a decrease is the physically consistent answer whenever the charge actually speeds up.