Problem 26 - University Physics with Modern Physics

Question

A particle has charge \(-3.00 \mathrm{nC}\) (a) Find the magnitude and direction of the electric field due to this particle at a point 0.250 \(\mathrm{m}\) directly above it. (b) At what distance from this particle does its electric field have a magnitude of 12.0 \(\mathrm{N} / \mathrm{C} ?\)

Step-by-Step Solution

Answer

(a) 431.52 N/C, downward; (b) 1.42 m.

Step 1: Understanding the Electric Field Concept

The electric field \( E \) due to a point charge \( q \) at a distance \( r \) from the charge is given by the formula \( E = \frac{k|q|}{r^2} \), where \( k = 8.99 \, \times 10^9 \, \mathrm{N} \, \mathrm{m}^2/\mathrm{C}^2 \) is Coulomb's constant.

Step 2: Calculate the Electric Field Magnitude at 0.250 m

Substitute the values into the electric field equation. Use \( q = -3.00 \, \mathrm{nC} = -3.00 \, \times \, 10^{-9} \, \mathrm{C} \) and \( r = 0.250 \, \mathrm{m} \). Thus, \( E = \frac{(8.99 \times 10^9 \, \mathrm{N} \, \mathrm{m}^2/\mathrm{C}^2)(3.00 \times 10^{-9} \, \mathrm{C})}{(0.250)^2} \). Calculating this gives \( E \approx 431.52 \, \mathrm{N}/\mathrm{C} \).

Step 3: Determine the Electric Field Direction

Since the charge is negative, the electric field direction is towards the charge. Therefore, at the point 0.250 m directly above the charge, the electric field direction is downward.

Step 4: Set up Equation for Given Electric Field Magnitude

We need to find the distance \( r \) where the electric field magnitude \( E \) is 12.0 \( \mathrm{N}/\mathrm{C} \). Rearrange the electric field equation to solve for \( r \): \( r = \sqrt{\frac{k|q|}{E}} \).

Step 5: Calculate the Distance for the Given Electric Field

Substituting the values \( E = 12.0 \, \mathrm{N}/\mathrm{C} \), \( q = 3.00 \, \times \, 10^{-9} \, \mathrm{C} \), and \( k = 8.99 \, \times \, 10^9 \, \mathrm{N} \, \mathrm{m}^2/\mathrm{C}^2 \) into the equation, we get \( r = \sqrt{\frac{(8.99 \times 10^9)(3.00 \times 10^{-9})}{12.0}} \approx 1.42 \, \mathrm{m} \).

Key Concepts

Coulomb's Law

Coulomb's Law is a fundamental principle in electrostatics. It describes how the electric force between two point charges is related to their magnitudes and the distance between them. This law states that the magnitude of the force is directly proportional to the product of the two electric charges and inversely proportional to the square of the distance between them.
The mathematical expression for Coulomb's Law is:

• \( F = \frac{k |q_1 q_2|}{r^2} \)
• Where:
• \( F \) is the magnitude of the force between the charges,
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges,
• \(r\) is the distance separating the charges,
• \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \; N \, m^2 / C^2 \).
The force acts along the line joining the two charges. If the charges have like signs, the force is repulsive; if the charges are of opposite signs, the force is attractive. In calculations involving a single charge and a point in space, the force translates into the electric field, which can be computed using similar principles.

Point Charge

A point charge is an idealized model of a charged object. It assumes that the charge is located at a single point in space, which simplifies calculations of electric fields and forces. This model is particularly useful when studying electrostatically charged particles, like electrons or protons, since they are small enough to be approximated as point charges.
In reality, charges are distributed over volumes, surfaces, or lines, but when the dimensions are negligible compared to the problem’s scope, the point charge model beautifully simplifies the equation for the electric field:

• The formula for the electric field \( E \) due to a point charge is \( E = \frac{k|q|}{r^2} \).
• \( q \) is the charge,
• \( r \) is the distance from the charge,
• \( k \) is Coulomb's constant.

This expression gives the electric field magnitude at any particular distance from the point charge. To apply this in a mathematical context, like the solved exercise, consider the particle's charge \( q = -3.00 \, nC\) and verify the direction using the charge's sign.

Electric Field Direction

The direction of an electric field is an equally important concept that goes hand-in-hand with the magnitude calculation. For a positive point charge, the electric field vectors emanate outward. For a negative charge, the field vectors point inward, directed toward the charge itself.
Understanding the direction gives insights into the nature of the electric field at a given point. Essentially, it guides you about how a positive test charge would move when placed in the field:

• For a negative point charge, like the one referenced in the exercise, the electric field at any point nearby will always be directed toward the charge itself.
• In the solved example, since the field point was directly above the charge and the charge was negative, the electric field direction was downward, indicating attraction toward the negative charge.

The direction is crucial for visualizing how the charge influences its surroundings and for understanding the resulting forces experienced by other charges in the field.