A point charge q moves from point P to point S along the path PQRS (Fig.) in a uniform electric fiel

Q: Why don't we calculate the work done for each segment (PQ, QR, RS) separately?

You absolutely can! But since the electrostatic force is conservative, the work done is path-independent. Calculating it directly from the initial to final position is a much faster and smarter shortcut.

Q: What if the electric field was not uniform?

If the electric field varied with position (e.g., $\vec{E} = x\hat{i}$), the force would not be constant. In that case, you would have to calculate the line integral $\int \vec{F} \cdot d\vec{r}$ along each specific segment of the path.

Q: Why is the work done negative?

The electric field points in the positive X-direction, so it pushes a positive charge to the right. However, the charge's overall displacement in the X-direction is from $x=a$ to $x=0$ (to the left). Since the force and displacement are in opposite directions, the work done is negative.

Q: Does the y-coordinate of the path affect the work done?

No. The electric field only has an X-component. Therefore, any movement in the Y-direction is perpendicular to the force, resulting in zero work done for that component of the displacement.

Q: What is the physical meaning of this negative work?

Negative work done by the electric field means the field is taking energy away from the charge. If the charge was moving freely, it would slow down. To move it along this path at a constant speed, an external force would have to do positive work.

Question

A point charge $q$ moves from point $P$ to point $S$ along the path $PQRS$ (Fig.) in a uniform electric field $E$ pointing parallel to the positive direction of the $X$-axis. The coordinates of points $P, Q, R$ and $S$ are $(a, b, 0), (2a, 0, 0), (a, -b, 0), (0, 0, 0)$ respectively. The work done by the field in the above process is given by the expression ...... .

Sigma Percentile · Accepted Answer

### Analyzing the Setup
Imagine the journey of our point charge $q$. It embarks on a multi-segment trip, starting at point $P(a, b, 0)$, traveling to $Q(2a, 0, 0)$, then to $R(a, -b, 0)$, and finally coming to rest at the origin $S(0, 0, 0)$. This entire journey takes place within a uniform electric field $\vec{E}$ that points steadily in the positive X-direction.

### The Master Equation
We are tasked with finding the total work done by the electric field on the charge. Here is where we can use a powerful shortcut. Because the electric field is uniform, the electrostatic force $\vec{F} = q\vec{E}$ acting on the charge is perfectly constant.

**A constant force is a conservative force.** This means the work it does is entirely path-independent! We don't need to calculate the work done along $PQ$, then $QR$, and then $RS$. We only care about the initial starting point and the final destination.

### Defining the Vectors
Let's write down our force and displacement vectors mathematically. The force vector is simply:

$\vec{F} = qE\hat{i}$

The total displacement vector $\vec{d}$ is the final position vector minus the initial position vector:

$\vec{d} = \vec{r}_S - \vec{r}_P$

Substituting the coordinates of $S$ and $P$:

$\vec{d} = (0\hat{i} + 0\hat{j}) - (a\hat{i} + b\hat{j})$

$\vec{d} = -a\hat{i} - b\hat{j}$

### Final Calculation
Now, we calculate the work done by taking the dot product of the force and displacement vectors:

$W = \vec{F} \cdot \vec{d}$

$W = (qE\hat{i}) \cdot (-a\hat{i} - b\hat{j})$

When taking the dot product, we multiply the corresponding components. Since the force only has an $\hat{i}$ component, the $\hat{j}$ component of the displacement does not contribute to the work done:

$W = (qE)(-a) + (0)(-b)$

**$W = -qEa$**

The negative sign is physically intuitive. The electric field pushes in the positive X-direction, but the charge's overall displacement in the X-direction was negative (from $x=a$ to $x=0$). Because the force and the effective displacement are in opposite directions, the work done by the field is negative.

Visualized Solution

Visualizing the Setup

The Power of Conservative Forces

Defining Force and Displacement

Calculating the Displacement Vector

Computing the Work Done

Final Answer

The Way Forward

The Sigma Insight: Electric Potential and Potential Difference

Analyzing the Setup Imagine the journey of our point charge $q$

The Master Equation We are tasked with finding the total work done by the electric field on the charge

Defining the Vectors Let's write down our force and displacement vectors mathematically

Final Calculation

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