Problem 19 - University Physics with Modern Physics

Question

Two point charges \(q_1 = +\)2.40 nC and \(q_2 = -\)6.50 nC are 0.100 m apart. Point \(A\) is midway between them; point \(B\) is 0.080 m from \(q_1\) and 0.060 m from \(q_2\) (\(\textbf{Fig. E23.19}\)). Take the electric potential to be zero at infinity. Find (a) the potential at point \(A\); (b) the potential at point \(B\); (c) the work done by the electric field on a charge of 2.50 nC that travels from point \(B\) to point \(A\).

Step-by-Step Solution

Answer

Potential at A is the sum of potentials from both charges at the midpoint. Potential at B considers each charge's distance at B. Work is the charge times the potential difference (A-B).

Step 1: Understand the Problem

We have two charges, \( q_1 = +2.40 \text{ nC} \) and \( q_2 = -6.50 \text{ nC} \), 0.100 m apart. We need to find the electric potential at two points: Point \( A \), which is halfway between the charges, and Point \( B \), which is 0.080 m from \( q_1 \) and 0.060 m from \( q_2 \). We also need to determine the work done by the electric field on a charge moving from \( B \) to \( A \). Electric potential should be considered zero at infinity.

Step 2: Calculate the Potential at Point A

Electric potential due to a point charge \( q \) is given by the formula \( V = \frac{kq}{r} \), where \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) is Coulomb's constant, and \( r \) is the distance from the charge. At midpoint \( A \), the distance \( r \) from both \( q_1 \) and \( q_2 \) is 0.050 m (half of 0.100 m). Calculate the potential contributions from both charges and add them:\[V_A = V_{q_1} + V_{q_2} = \frac{k q_1}{0.050} + \frac{k q_2}{0.050}\]Substituting in:\[V_A = \frac{8.99 \times 10^9 \times 2.40 \times 10^{-9}}{0.050} + \frac{8.99 \times 10^9 \times (-6.50) \times 10^{-9}}{0.050}\]Calculate each term and then combine them to find \( V_A \).

Step 3: Calculate the Potential at Point B

For Point \( B \), use the same formula, but with the distances given for each charge's influence at \( B \): 0.080 m for \( q_1 \) and 0.060 m for \( q_2 \):\[V_B = \frac{k q_1}{0.080} + \frac{k q_2}{0.060}\]Substituting in:\[V_B = \frac{8.99 \times 10^9 \times 2.40 \times 10^{-9}}{0.080} + \frac{8.99 \times 10^9 \times (-6.50) \times 10^{-9}}{0.060}\]Calculate each term and combine them to obtain \( V_B \).

Step 4: Calculate the Work Done Moving from B to A

The work done by the electric field moving a charge \( q = 2.50 \text{ nC} \) from point \( B \) to point \( A \) is calculated using the change in electric potential: \( W = q \cdot (V_A - V_B) \).First, find the potential difference:\[ \Delta V = V_A - V_B \]Then use:\[W = (2.50 \times 10^{-9}) \times \Delta V\]Substitute the calculated potentials from Step 2 and Step 3 to find the work done.

Key Concepts

Electric Field

The electric field is a vector field surrounding an electric charge that exerts a force on other charges, attracting or repelling them. The direction of the electric field is always in the direction a positive test charge would move if placed within the field. To visualize, imagine lines radiating outward from a positive charge, each one indicating the force vector. The density of these lines represents the field's strength. As you move further from the charge, the electric field weakens.
To calculate the electric field (E) produced by a point charge (q), we use the formula: \[ E = \frac{k |q|}{r^2} \]where \( E \)is the magnitude of the electric field, \( r \) is the distance from the charge, and \( k \) is Coulomb's constant. The force experienced by another charge \( q_0 \) placed a certain distance from the original point charge is given by \[ F = q_0 \times E \]. This formula is incredibly useful in determining how a charged particle will behave in the presence of others, deriving directly from Coulomb's law.

Point Charges

Point charges are idealized charges that are infinitely small yet possess a significant amount of charge value. In the real world, no charge can truly be a point, but this model simplifies calculations and theoretical physics studies.
In exercises involving point charges, like the one described, we assume each charge is at a precise location with no physical dimensions. This allows us to use formulas for electric fields and potentials more straightforwardly, as the geometry of the system is simplified.

• For a single point charge, the electric potential (V) at a certain distance (r) can be calculated using: \[ V = \frac{kq}{r} \]where \( k \) is Coulomb's constant, and \( q \) is the charge's magnitude.
• In a system with multiple charges, the superposition principle helps determine the total potential or field, calculated by summing contributions from all individual charges.

This principle implies that each charge impacts the surroundings independently, making it easier to calculate complex systems' overall effects by adding individual contributions.

Coulomb's Constant

Coulomb's constant, denoted as \( k \), is a fundamental value used in various equations describing electric forces between charges. It derives from Coulomb's Law, which states that the force (\( F \)) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.
Coulomb's constant quantifies this relationship, with a standard value of \( k = 8.99 imes 10^{9} ext{ Nm}^2/ ext{C}^2 \). This constant helps ensure that our units for charge, distance, and force are consistent, enabling accurate calculations in electrostatics problems.

• It's essential when using formulas for electric potential and fields, like \( V = \frac{kq}{r} \)and \( E = \frac{k |q|}{r^2} \), respectively.
• The constant simplifies to a value convenient for calculations when the charges are measured in coulombs and the distance in meters, a standard in physics.

Understanding \( k \) is crucial in solving problems involving electric interactions, as it connects theoretical predictions with measurable phenomena.