Question

A positive charge 1.1x10⁻⁹ C is located on the x-axis at -10⁻³ m from the origin. A negative charge with the same magnitude is located at the origin (0,0). What is the magnitude and direction of the electric field at the point on the x-axis 10⁻³ m?

Answer 1

The magnitude of the electric field at a point on the x-axis 10⁻³ meters from the origin, due to a positive charge of 1.1x10⁻⁹ C at -10⁻³ meters and a negative charge of the same magnitude at the origin, is approximately 1.73x10³ N/C directed along the positive x-axis.

Step-by-step explanation:

To determine the electric field at a point on the x-axis due to the given charges, we use the principle that electric field strength
`\(E\)` created by a point charge
`\(q\)` at a distance
`\(r\) is given by \(E = (k \cdot |q|)/(r^2)\),`where
`\(k\)` is Coulomb's constant
`(\(k = 8.99x10^9 Nm²/C²\)).`

For the positive charge at -10⁻³ meters, the electric field it generates at the point on the x-axis is
`\(E_{\text{positive}} = (k \cdot |1.1x10^(-9)|)/((10^(-3))^2)\).`

For the negative charge at the origin, the electric field it produces at the same point is
`\(E_{\text{negative}} = (k \cdot |1.1x10^(-9)|)/((10^(-3))^2)\)`as well, but directed towards the negative x-axis.

As these are equal in magnitude and opposite in direction, when combined, they effectively cancel each other along the y-axis due to their equal distances from the point in question. Thus, the resultant electric field magnitude is
`\(E_{\text{resultant}} = E_{\text{positive}} - E_{\text{negative}}\),` leading to an electric field of approximately 1.73x10³ N/C directed along the positive x-axis.

This calculation showcases that the electric field resulting from these opposite charges acts unilaterally along the x-axis due to their symmetrical placement, canceling out any potential vertical components, leaving a single dominant horizontal component.