Problem 63 - A Complete Resource Book in Physics for JEE Main

Question

A point charge \(q\) and a charge \(-q\) are placed at \(x=-a\) and \(x=+a\), respectively. Which of the following represents a part of \(E-x\) graph?

Step-by-Step Solution

Answer

The net electric field \(E_x\) at position x on the x-axis due to two point charges, positive charge q at x = -a and negative charge -q at x = +a, is given by: \[E_x = k \frac{q}{(-a - x)^2} - k \frac{q}{(a - x)^2}\] The behavior of \(E_x\) with respect to x can be summarized as follows: 1. When x is between -a and +a, \(E_x\) is positive and maximum at x = 0. 2. When x < -a, \(E_x\) is negative and decreasing towards the positive charge. 3. When x > a, \(E_x\) is positive and increasing towards the positive charge.

Step 1: Identify the charges and their locations

We are given two point charges: 1. A positive point charge q located at x = -a. 2. A negative point charge -q located at x = +a.

Step 2: Find the electric field due to each charge using Coulomb's Law

Coulomb's Law states that the electric field E due to a point charge Q is given by the following equation: \[E = \frac{kQ}{r^2}\] where: - E is the electric field, - k is Coulomb's constant = \(8.99 \times 10^9 Nm^2 C^{-2}\), - Q is the point charge, - r is the distance between the point charge and the point where we want to measure the electric field. For each charge at a position x on the x-axis, we can calculate the electric fields: 1. Electric field due to positive charge q at position x: \[E_{pos} = k \frac{q}{(-a - x)^2}\] 2. Electric field due to negative charge -q at position x: \[E_{neg} = -k \frac{q}{(a - x)^2}\]

Step 3: Determine the net electric field at position x on the x-axis

To determine the net electric field, we'll add the two electric fields from the previous step: \[E_x = E_{pos} + E_{neg}\] Substituting the values for E_pos and E_neg: \[E_x = k \frac{q}{(-a - x)^2} - k \frac{q}{(a - x)^2}\]

Step 4: Analyze how the electric field behaves with respect to x

Now we will analyze the behavior of the net electric field Ex with respect to the position x on the x-axis: 1. When x is between -a and +a (i.e., x < a and x > -a), the electric field due to the positive charge and negative charge will be in the same direction (both fields push towards the positive charge). So, the net electric field Ex will be positive and maximum at x = 0. 2. When x is less than -a (i.e., x < -a), the electric field due to the positive charge will be stronger than the electric field due to the negative charge (since the distance to the positive charge is less). As a result, the net electric field Ex will be negative and decreasing towards the positive charge. 3. When x is greater than +a (i.e., x > a), the electric field due to the negative charge will be stronger than the electric field due to the positive charge (since the distance to the negative charge is less). In this case, the net electric field Ex will be positive and increasing towards the positive charge. From this analysis, we can create a part of the Ex graph that shows electric field behavior along the x-axis.