91Ó°ÊÓ

An electric field is a vector field that represents the force exerted per unit charge at any point in space. In our exercise, the electric field is described by the function \( \vec{E} = \vec{i} A x \), where \( A = 10 \, \text{V/m}^2 \), indicating that the field's strength changes linearly along the \( x \)-axis. Typically, electric fields exert forces on charged particles, driving them along specific directions.
This electric field has a unique property - it only has a component in the \( x \)-direction (noted with \( \vec{i} \)). Understanding how the field changes with \( x \) is crucial for determining other physical quantities.
This field implies that each metre displacement in the \( x \)-axis results in a change in the field strength, which is indicative of a linearly varying electric field. Simply put, as you move farther away from the origin along the \( x \)-axis, the field strength increases proportionally.

Potential difference, often called voltage, is the difference in electric potential energy between two points in an electric field. Imagine it as the energy needed to move a charge from one point to another within an electric field.
In our problem, potential difference is related to the electric field through the relation \( E = -\frac{dV}{dx} \). This formula highlights that the electric field can be seen as the spatial rate of change of electric potential. Negative sign here indicates the potential decreases in the direction of the electric field.
For this exercise, knowing the potential difference at a specific point, i.e., \((10, 20)\) is zero, helps us calculate the integration constant needed to derive potential at other points, such as the origin. This setup allows us to establish a reference point for measurements, making it easier to find potentials at other key locations.

Integration in the context of an electric field often involves finding the potential function from a known electric field function. This process essentially reverses differentiation, allowing us to compute the potential at any point based on the field's characteristics.
To go from \( E(x) = Ax \) to \( V(x) \), we integrate \( E \). The integral \( V(x) = -A \int x \, dx \) becomes \( V(x) = -\frac{Ax^2}{2} + C \), where \( C \) is an integration constant.

• The integral gives us a family of functions, all differing by a constant, \( C \).
• The constant \( C \) is determined using initial or boundary conditions, like knowing \( V(10, 20) = 0 \).

Substituting these conditions helps us solve for \( C \) and gives us a fully defined potential function. In this problem, calculating the potential difference and knowing its value at certain points are key to solving for \( C \) and finding potential at the origin (\( 0, 0 \)).