Problem 24 - University Physics with Modern Physics

Question

At a certain distance from a point charge, the potential and electric-field magnitude due to that charge are 4.98 \(\mathrm{V}\) and \(12.0 \mathrm{V} / \mathrm{m},\) respectively. (Take the potential to be zero at infinity.) (a) What is the distance to the point charge? (b) What is the magnitude of the charge? (c) Is the electric field directed toward or away from the point charge?

Step-by-Step Solution

Answer

(a) 0.415 m, (b) 2.30 × 10^{-10} C, (c) Away from the charge.

Step 1: Understanding the Relationship

The electric field magnitude \( E \) and the electric potential \( V \) due to a point charge \( q \) at a distance \( r \) from the charge are given by the formulas: \( E = \frac{k \cdot q}{r^2} \) and \( V = \frac{k \cdot q}{r} \), where \( k \) is Coulomb's constant (\(8.99 \times 10^9 \; \text{N} \cdot \text{m}^2/\text{C}^2\)).

Step 2: Setting Up Equations for Distance

From the given formulas, for the potential, \( V = \frac{kq}{r} = 4.98 \; \text{V} \), and for the electric field, \( E = \frac{kq}{r^2} = 12.0 \; \text{V/m} \). These equations will help us to solve for \( r \).

Step 3: Deriving the Expression for Distance

By dividing the potential equation by the electric field equation, we get: \( \frac{kq}{r} \div \frac{kq}{r^2} = 4.98 \div 12.0 \), which simplifies to \( r = \frac{4.98}{12.0} \approx 0.415 \; \text{m} \). This is the distance to the point charge.

Step 4: Calculate Charge Magnitude

We have \( V = \frac{kq}{r} = 4.98 \; \text{V} \). Solving for \( q \), we rearrange to \( q = \frac{Vr}{k} = \frac{4.98 \cdot 0.415}{8.99 \times 10^9} \approx 2.30 \times 10^{-10} \; \text{C} \). This is the magnitude of the charge.

Step 5: Analyzing Electric Field Direction

The electric field due to a positive point charge radiates away from the charge, while for a negative charge, it points towards it. Since the problem doesn't specify the sign of the charge, and electric field direction is typically away from positive charges, we assume the charge is positive, so the electric field is directed away from the point charge.

Key Concepts

Electric Field

The electric field is a fundamental concept in physics, particularly in electromagnetism. It represents the force per unit charge exerted by an electric charge on another charged object in its vicinity. The strength and direction of this force are described by the electric field vector, which is defined as:

• F: The force experienced by a small positive test charge.
• q: The magnitude of the test charge.

The formula for the electric field, denoted as \( E \), due to a point charge is:\[ E = \frac{k \, q}{r^2} \]where:

• \( k \) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \)
• \( q \) is the charge of the point source creating the field
• \( r \) is the distance from the charge to the point where the field is being calculated

Understanding how the electric field works can help in predicting how charges will interact with one another. The field's direction is always oriented from positive to negative charges.

Point Charge

A point charge is an idealized model of a charge that is concentrated at a single point in space. This simplification allows for easier calculations in theoretical physics and is often used as an approximation in electromagnetic studies. Here are some important points about point charges:

• They are fundamental sources of electric fields.
• They can either be positive or negative, which influences the direction of their electric field.
• The electric field and potential around a point charge decrease with the square of the distance from the charge (as per Coulomb's law).
• Despite their use in theory, real charges are not actually point charges but have physical sizes and shapes.

In the context of this exercise, the point charge creates an electric potential and field at a certain distance, which allows us to calculate particular properties such as distance and charge magnitude.

Coulomb's Constant

Coulomb's constant is a key figure in the study of electric forces between point charges. It appears in Coulomb's Law, which provides a formula for the force between two charges and is an essential parameter in determining electric fields and potentials. Here is more on Coulomb's constant:

• It is denoted as \( k \).
• The value of \( k \) is approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \).
• This constant helps relate the distance between the charges and the magnitude of the force or field.
• In calculations, \( k \) allows transformations between force and electric field strength for point charges or systems of charges.

In practice, understanding Coulomb's constant contributes significantly to mastery in physics, especially in problems relating to electrostatics where point charges interact.