Problem 43 - Arihant AIEEE Physics

Question

An electrostatic field \(E\) of magnitude \(10 \mathrm{~N} / \mathrm{C}\) is directed along positive \(x\) -axis. A point charge of \(10^{-6} \mathrm{C}\) is shifted from \(A(1 \mathrm{~m}, 0)\) to \(B(2 \mathrm{~m}, 0,1 \mathrm{~m})\), then from point \(B\) to \(C(0,0,0)\), the work done by electrostatic force is: (a) \(-10^{-5} \mathrm{~J}\) (b) \(10^{-5} \mathrm{~J}\) (c) \(-10^{-4} \mathrm{~J}\) (d) none of these

Step-by-Step Solution

Answer

The work done is \(10^{-5} \text{ J}\), option (b).

Step 1: Calculate Work Done from A to B

The work done by an electric field along a path is given by the formula:\[ W = -q \int_{r_1}^{r_2} \mathbf{E} \cdot d\mathbf{r} \]where \( q \) is the charge, \( \mathbf{E} \) is the electric field, and \( d\mathbf{r} \) is the displacement vector.For a uniform electric field directed along the x-axis, \[ \mathbf{E} = 10 \hat{i} \text{ N/C} \].The displacement from \( A(1,0,0) \) to \( B(2,0,1) \) is \[ \mathbf{r} = (2-1)\hat{i} + (0-0)\hat{j} + (1-0)\hat{k} = 1 \hat{i} + 0 \hat{j} + 1 \hat{k} \text{ m} \].Therefore, the work done is:\[ W_{AB} = -q \mathbf{E} \cdot \mathbf{r} = -(10^{-6} C)(10 \hat{i}) \cdot (1 \hat{i} + 0 \hat{j} + 1 \hat{k}) \]\[ W_{AB} = -10^{-6} \times 10 \times (1) = -10^{-5} \text{ J} \]

Step 2: Calculate Work Done from B to C

The displacement from \( B(2,0,1) \) to \( C(0,0,0) \) is \[ \mathbf{r} = (0-2)\hat{i} + (0-0)\hat{j} + (0-1)\hat{k} = -2 \hat{i} - 1 \hat{k} \text{ m} \].The work done is:\[ W_{BC} = -q \mathbf{E} \cdot \mathbf{r} = -(10^{-6} C)(10 \hat{i}) \cdot (-2 \hat{i} - 1 \hat{k}) \]\[ W_{BC} = -10^{-6} \times 10 \times (-2) = 2 \times 10^{-5} \text{ J} \]

Step 3: Calculate Total Work Done

The total work done is the sum of the work done over the two paths:\[ W_{total} = W_{AB} + W_{BC} = -10^{-5} + 2 \times 10^{-5} = 10^{-5} \text{ J} \]

Step 4: Determine the Correct Option

Based on the calculation, the total work done by the electrostatic force is \(10^{-5} \text{ J} \), so the correct answer is option (b).

Key Concepts

Work Done by Electric Field

In electrostatics, work done by the electric field is crucial to understand energy transformations. When a charge is moved in an electric field, the field does or requires work, depending on the path and direction. The formula for the work done by an electric field is given by:

• \( W = -q \int_{r_1}^{r_2} \mathbf{E} \cdot d\mathbf{r} \)

where \( q \) is the charge, \( \mathbf{E} \) is the electric field vector, and \( d\mathbf{r} \) is the displacement vector. The negative sign indicates that work done by the electric field reduces the system's potential energy.
To calculate the work done, we must evaluate the dot product of the electric field and the displacement vector, which highlights the components of movement along the direction of the electric field. Only the parallel components contribute to the work, as perpendicular movements do not require work in uniform fields.

Displacement Vector

The displacement vector is a fundamental component to determine the path taken by a charge in an electric field. It represents the movement from an initial point to a final point in vector terms. In our example, the displacement vector from point \( A(1,0,0) \) to point \( B(2,0,1) \) is calculated by subtracting the coordinates of point A from those of point B:

• \( \mathbf{r} = (2-1)\hat{i} + (0-0)\hat{j} + (1-0)\hat{k} = 1 \hat{i} + 0 \hat{j} + 1 \hat{k} \text{ m} \)

This vector indicates the distance moved in each dimension and helps determine which portions of the path contribute to work done by the electric field. The calculations must carefully adhere to direction signs, as they affect calculations in the integral required to compute work.

Uniform Electric Field

A uniform electric field has the same strength and direction at every point, simplifying calculations significantly. Such fields are often described by a constant vector. For instance, if the electric field vector \( \mathbf{E} \) is \( 10 \hat{i} \text{ N/C} \), it uniformly points along the positive x-axis.
In a uniform field, the work done is directly related to the linear displacement along the field lines, making calculations easier. This means we can use simple vector products without needing complex calculus, as the field does not vary in space. It implies that only displacements parallel to the direction of the field contribute to the work done. Hence, perpendicular movements are energy neutral.

Point Charge Movement

The movement of a point charge in an electric field involves understanding forces and energy changes due to electrostatic interactions. When a charge is moved, it experiences a force equal to

• \( \mathbf{F} = q \mathbf{E} \)

This force tends to align the charge's path with the field direction, influencing work calculations.
Also, analyzing a point charge’s path reveals how different portions of the trajectory contribute to the total work done. Starting at \( A(1,0,0) \), moving to \( B(2,0,1) \), then to \( C(0,0,0) \) provides ample insight into directional dependence of work in fields.
By looking at displacements, the charge first moves in the positive x direction, contributing negatively to work when returning along axis directions. This intricately connects the concepts of electrostatic force and energy with movement in space.