Problem 51 - University Physics with Modern Physics

Question

A ring-shaped conductor with radius \(a =\) 2.50 cm has a total positive charge \(Q = +\)0.125 nC uniformly distributed around it (see Fig. 21.23). The center of the ring is at the origin of coordinates \(O\). (a) What is the electric field (magnitude and direction) at point \(P\), which is on the \(x\)-axis at \(x =\) 40.0 cm? (b) A point charge \(q = -2.50 \space \mu\)C is placed at \(P\). What are the magnitude and direction of the force exerted by the charge \(q\) \(on\) the ring?

Step-by-Step Solution

Answer

The electric field at P is approximately 6.988 N/C, directed along the x-axis. The force on the point charge is approximately -17.47 mN, directed towards the ring.

Step 1: Understand the Setup

We have a ring with a uniform positive charge distribution and we are interested in finding the electric field at point P on the x-axis. The point P is 40.0 cm from the origin, which is the center of the ring. The radius of the ring is 2.50 cm. The ring's charge is positive.

Step 2: Determine the Electric Field at Point P

Since the electric field is being calculated at a point on the axis of the ring, we use the formula for the electric field due to a ring: \[E = \frac{k_e Q x}{(x^2 + a^2)^{3/2}}\]where \(k_e = 8.99 \times 10^9 \text{ N m}^2/\text{C}^2\), \(Q = 0.125 \text{ nC} = 0.125 \times 10^{-9} \text{ C}\), \(a = 0.025 \text{ m}\), and \(x = 0.4 \text{ m}\). Substitute these values into the formula to find the magnitude of the electric field.

Step 3: Calculate the Magnitude of the Electric Field

Plug in the values into the formula:\[E = \frac{(8.99 \times 10^9) (0.125 \times 10^{-9}) (0.4)}{(0.4^2 + 0.025^2)^{3/2}}\]Calculate: \(E \approx \frac{(8.99 \times 10^9)(0.125 \times 10^{-9})(0.4)}{(0.160625)^{3/2}}\)\(E \approx \frac{0.4495}{0.0643} \approx 6.988 \text{ N/C}\).

Key Concepts

Coulomb's Law

Coulomb's Law is fundamental in understanding electric forces between charged objects. It states that the electric force (F) between two point charges is proportional to the product of their charges and inversely proportional to the square of the distance between them.- Mathematically, it is expressed as:\(F = \frac{k_e |q_1 q_2|}{r^2}\)- Here, \(k_e\) is Coulomb's constant (8.99 \times 10^9 \space \text{N m}^2/\text{C}^2), \(q_1\) and \(q_2\) are the charges, and \(r\) is the distance between them.In this exercise, although we deal with a continuous distribution of charge around a ring, the principles of Coulomb's law still apply. Each infinitesimally small charge element on the ring contributes to the electric field at point \(P\). Summing these contributions using Coulomb's Law leads to the calculation of the total electric field at that point.

Electric Forces

Electric forces are pivotal in physics, and they arise due to the interaction between charged objects. In this scenario, we are interested in the electric force exerted by a charged ring on another charge placed at point \(P\).- The ring is uniformly charged with a total charge \(Q = +0.125 \space \text{nC}\).- We consider the electric field created by the ring at point \(P\) using the formula for the field of a charged ring.After determining the electric field at \(P\), we can calculate the force exerted on a point charge \(q = -2.50 \space \mu\text{C}\) placed at \(P\) using:- \(F = qE\)- Here, \(E\) is the electric field calculated using the distribution of the ring's charge.Because the charge \(q\) is negative, the direction of the electric force will be opposite to the direction of the electric field.

Charge Distribution

Understanding charge distribution is key for analyzing electric fields from objects with complex shapes. Here, the ring has a uniform charge distribution, meaning the charge is evenly spread out along the ring's circumference.- The total charge of the ring is \(Q = +0.125 \space \text{nC}\).- Due to this distribution, each infinitesimal segment of the ring contributes evenly to the electric field at distant points.To find the electric field at point \(P\), we integrate the contributions of all these small charge segments. This approach simplifies the complexity of dealing with countless point charges, allowing us to use symmetry and calculus to derive the field at any point along the axis of the ring.

Physics Education

In physics education, understanding how to calculate the electric field due to various charge distributions is critical. This exercise demonstrates how integral calculus and symmetry play a role in simplifying and solving problems involving distributed charges. - Grasping the concept of charge distributions helps in solving more complex scenarios beyond simple point charges. - Exercises like this one train students in applying mathematical tools to real-world physics problems, which is an essential skill. By engaging with such problems, students develop a deeper understanding of electric forces and the mathematics behind them, preparing them for more advanced topics in electrostatics and electromagnetism.