Problem 22 - 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 V and 16.2 V\(/\)m, respectively. (Take \(V = 0\) 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) The distance is 0.307 m. (b) The magnitude of the charge is approximately \(1.7 \times 10^{-10}\) C. (c) The electric field is directed away from the charge.

Step 1: Understanding the Relationships

Understand that the electric potential \( V \) and electric field \( E \) for a point charge \( Q \) at a distance \( r \) are given by \( V = \frac{kQ}{r} \) and \( E = \frac{kQ}{r^2} \) respectively, where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \)). The goal is to find \( r \) and \( Q \).

Step 2: Finding the Distance to the Point Charge

By dividing the formula for \( V \) by the formula for \( E \), we get: \( \frac{V}{E} = r \). Plug the known values into this equation: \( \frac{4.98}{16.2} = r \), yielding \( r = 0.307 \) m.

Step 3: Calculating the Magnitude of the Charge

Using the value of \( r = 0.307 \) m and the formula for \( V \) (\( V = \frac{kQ}{r} \)), we solve for \( Q \). Rearrange to find \( Q = \frac{Vr}{k} \). Substituting the known values, \( Q = \frac{4.98 \times 0.307}{8.99 \times 10^9} \), resulting in \( Q \approx 1.7 \times 10^{-10} \) C.

Step 4: Determining the Direction of the Electric Field

Since the electric field magnitude \( E \) and potential \( V \) are both positive, the point charge \( Q \) is positive. The electric field is directed away from positive charges.

Key Concepts

Electric Potential

Electric potential is a scalar quantity that is associated with the electric field a charge produces. It is the amount of work needed to move a unit positive charge from a reference point (typically at infinity) to a specific point in the space around a charge, without any acceleration. This is expressed as the potential energy per unit charge.
For a point charge, the electric potential at a distance \( r \) from the charge \( Q \) is given by the formula:

• \( V = \frac{kQ}{r} \)

where \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2 \).
In the original problem, the potential \( V \) given is 4.98 V, suggesting the work done per unit charge to move it from infinity to the current position is 4.98 joules.
This scalar nature means that unlike vector quantities, potentials can be added algebraically without considering direction, making calculations straightforward.
Understanding electric potential is key to solving many electrostatics problems, especially those involving energy considerations.

Electric Field

The electric field is a vector field that represents the force experienced by a test charge at any point in space due to other charge distributions. It tells us the magnitude and direction of the force that would be felt by a unit positive charge placed at any given point.
For a point charge, the electric field \( E \) can be calculated using the equation:

• \( E = \frac{kQ}{r^2} \)

An electric field magnitude of 16.2 V/m indicates the force per unit charge at the given distance from the charge is 16.2 newtons per coulomb.
It's vectorial nature makes the electric field more complex to deal with compared to scalar quantities like electric potential. It involves directional considerations. For example, the electric field direction is determined by the nature of the source charge; it points away from positive charges and towards negative ones.
Grasping the concept of electric fields aids in visualizing how charges influence their surroundings and helps solve problems related to forces and motion under electric influences.

Coulomb's Law

Coulomb's law is the foundational principle that describes how charged particles interact with one another. It mathematically defines the electric force between two point charges.
The law is represented by the equation:

• \( F = \frac{k|Q_1 Q_2|}{r^2} \)

Here, \( F \) is the force between charges \( Q_1 \) and \( Q_2 \), \( r \) is the distance between the charges, and \( k \) is Coulomb's constant.
This concept is crucial for understanding electric interactions. It tells us:

• The force is inversely proportional to the square of the distance, meaning it decreases rapidly with distance.
• The force's magnitude is proportional to the product of the magnitudes of the charges involved.
• Opposite charges attract, so the force is negative, while like charges repel, resulting in a positive force.

In the given problem, Coulomb's Law helps determine the direction and magnitude of the forces and fields created by point charges, providing insights into how charges behave and influence one another in electrostatic situations.