91Ó°ÊÓ

To understand where the electric field is zero or in a specific direction, we must first understand how to calculate the electric field due to a point charge. The electric field, represented by the symbol \(E\), is generated by a charge \(q\) and is influenced by the distance \(r\) away from the charge. The formula we use for the electric field is \(E = \frac{kq}{r^2}\), where \(k\) is Coulomb's constant. This equation tells us how strong the electric field is at a certain distance from the charge.

• The unit of charge is esu in this problem.
• Positive charges create an electric field pointing away from the charge.
• Negative charges create an electric field pointing towards the charge.

For two charges along an axis, the total electric field at a specific point is the sum of the fields produced by each charge. When we say an electric field is zero at a point, it means the fields due to individual charges cancel each other perfectly at that point.

A point charge refers to an idealized model of a charge that is assumed to be located at a single point in space. In practice, real charges are distributed over a space, but for simplicity, especially in theoretical problems, point charges help us visualize the problem.
In this exercise, we deal with:

• A +1 esu charge located at the origin (0,0).
• A -2 esu charge located at \(x = 1\) on the x-axis.

These charges are on a line (the x-axis), and we must determine how they affect the space around them, specifically where one field cancels the other. It's essential to understand that the electric field direction is as much about geometry as it is about charge magnitude.

In this exercise, finding the point where the electric field is zero involved solving a quadratic equation. Quadratic equations are common in electrostatics calculations, especially when the fields from two charges have to be balanced. The process involves setting up the balance of fields equation and simplifying it to a form we can solve:
The equation from the exercise refers to balancing the fields from two charges:\[\frac{1}{x^2} = \frac{2}{(1-x)^2}\]Solving for \(x\) using algebra, we cross-multiply and simplify to get:\[x^2 = 2(1 - 2x + x^2)\]This simplifies to:\[0 = x^2 - 4x + 2\]Using the quadratic formula \(x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\), where \(a = 1\), \(b = -4\), and \(c = 2\), the solutions become \(x = 2 \pm \sqrt{2}\). Such manipulation is vital for finding exact points of field interactions, offering a mathematically structured way to pinpoint solutions within electrostatic problems.