91Ó°ÊÓ

The electric field is a fundamental concept in electrostatics. It is a vector quantity, meaning it has both magnitude and direction. The electric field describes the force exerted per unit charge on a test charge placed in the vicinity of another charge. Imagine it as the "aura" or "influence zone" around a charged object.
In the context of our problem, we are concerned with the electric field just outside the surface of a spherical conductor. This electric field is generated by the charge distributed over the sphere's surface. In general, electric fields show us how a charged object will interact with other charged objects nearby.

To calculate the electric field due to a charge, we use the formula:

• \[ E = \frac{F}{q} \]
• where: \( E \) is the electric field, \( F \) is the force, and \( q \) is the test charge.

This formula simplifies to

• \[ E = \frac{Q}{4\pi\varepsilon_0r^2} \]
• where \( Q \) is the charge on the conductor, \( \varepsilon_0 \) is the permittivity of free space, and \( r \) is the radius of the sphere.

By using this relationship, we can determine how strong the electric field will be given a certain distribution of charge.

To find out how many excess electrons need to be added to a conductor to create a specific electric field, we must first calculate the necessary charge. This involves rearranging the formula for electric field designed for spherical conductors.
The formula \( E = \frac{Q}{4\pi\varepsilon_0r^2} \) can be rearranged to solve for the charge \( Q \):

• \[ Q = E \cdot 4\pi\varepsilon_0r^2 \]

By substituting the given electric field and known constants into this equation, we compute \( Q \), the charge that must be placed on the conductor.
In our case, substituting the given values yields:

• \[ Q = 1150 \times 4\pi \times 8.85 \times 10^{-12} \times (0.16)^2 \]
• Simplifying, we find \( Q \approx 5.18 \times 10^{-9} \text{ C} \).

Once we have the charge \( Q \), we can easily convert it into the number of electrons, knowing the charge of one electron is \( e = 1.60 \times 10^{-19} \text{ C} \). The number of excess electrons needed is thus:

• \[ n = \frac{Q}{e} = \frac{5.18 \times 10^{-9}}{1.60 \times 10^{-19}} \]
• \[ n \approx 3.24 \times 10^{10} \]

This calculation helps us understand the direct relationship between charge, electric field, and the number of electrons.

The permittivity of free space, denoted as \( \varepsilon_0 \), is a key constant in electrostatics equations. It quantifies the ability of the vacuum of space to permit electric field lines. To put it simply, it affects how a charge will "spread out" in space.
The permittivity of free space is used in many standard equations relevant to electrical interactions and forces, like the one for the electric field, Coulomb's Law, and capacitance calculations.
In the electric field formula, it appears as:

• \[ E = \frac{Q}{4\pi\varepsilon_0 r^2} \]
• where \( \varepsilon_0 \) is approximately \( 8.85 \times 10^{-12} \text{ C}^2/\text{N}\cdot\text{m}^2 \).

This constant is crucial in calculating how electric fields behave in a vacuum and helps us predict the force between charged objects.
Understanding \( \varepsilon_0 \) is important because it sets the baseline for how electric fields behave and interact in a world without any other interfering materials. Its role in electrostatics is vast as it establishes the foundational behavior of electric forces in most classical physics models.