Question

A point charge \(q\) is situated at a distance \(r\) from one end of a thin conducting rod of length \(\mathrm{L}\) having a charge \(\mathrm{Q}\) (uniformly distributed along its length). The magnitude of electric force between the two, is ...... \((\mathrm{A})[(2 \mathrm{kqQ}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\) (B) \([(\mathrm{kq} \mathrm{Q}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\) (C) \([(\mathrm{kqQ}) / \mathrm{r}(\mathrm{r}-\mathrm{L})]\) (D) \([(\mathrm{kQ}) / \mathrm{r}(\mathrm{r}+\mathrm{L})]\)

Short answer

The magnitude of the electric force between the point charge and conducting rod is: \(\boxed{(B) \frac{kqQ}{r(r+L)}}\)

Step-by-step solution

Identify the given quantities and relevant equations

We are given the following quantities: - Point charge: q - Distance from end of the rod: r - Length of the conducting rod: L - Charge of the conducting rod: Q (uniformly distributed) The relevant equations for this problem are: 1. Coulomb's Law: \(F = \frac{kq_1q_2}{r^2}\), where F is the force between two charges, k is the Coulomb's constant (\(8.99 x 10^9 Nm^2C^{-2}\)), \(q_1\) and \(q_2\) are the charges, and r is the distance between them. 2. Charge density \(\lambda=\frac{Q}{L}\).

Set up the infinitesimal charge element and the electric force between the charge element and the point charge

We'll consider an infinitesimal charge element dq at a distance x from the end of the rod. Since the charge Q is uniformly distributed, the charge density is constant along the rod's length. The charge density is given by \(\lambda=\frac{Q}{L}\). Thus, the infinitesimal charge element at a distance x from the end of the rod is given by dq = \(\lambda dx = \frac{Q}{L}dx\). Now let's find the infinitesimal electric force between the point charge q and the infinitesimal charge element dq. Using Coulomb's law, the force df is given by: \(df = \frac{kqdq}{(r+x)^2}\) Substitute dq from above: \(df = \frac{kq(Q/L)dx}{(r+x)^2}\)

Integrate the infinitesimal force over the length of the rod

To find the total electric force between point charge q and the conducting rod, we need to integrate the infinitesimal force df over the entire length of the rod (0 to L): \(F = \int_0^L df = \int_0^L \frac{kq(Q/L)dx}{(r+x)^2}\) Now, let's solve this integral: \(F = \frac{kqQ}{L} \int_0^L \frac{dx}{(r+x)^2}\) To solve this integral, we can use the substitution method. Let \(u = r + x\), so the differential \(du = dx\). Also, when x = 0, \(u = r\), and when x = L, \(u = r + L\). Now we can rewrite and solve the integral in terms of u: \(F = \frac{kqQ}{L} \int_r^{r+L} \frac{du}{u^2} = -\frac{kqQ}{L} \left[ \frac{1}{u} \right]_r^{r+L}\)

Evaluate the integral and find the magnitude of the electric force

Now, let's substitute the limits of the integral and find the magnitude of the electric force: \(F = \frac{kqQ}{L} \left( -\frac{1}{r + L} + \frac{1}{r} \right)\) Rearrange the terms: \(F = \frac{kqQ}{r(r + L)}\) Comparing to the given options, we find that the magnitude of the electric force between the point charge and conducting rod is: \(\boxed{(B) \frac{kqQ}{r(r+L)}}\)

Key concepts

Electric Force

The concept of electric force is rooted in Coulomb's Law, which is fundamental in physics for describing interactions between charged objects. The electric force is a vectored force, meaning it has both magnitude and direction, that acts between charged particles. It is a non-contact force, meaning the charged objects do not have to be touching to exert force on one another.

Coulomb's Law provides a formula to calculate this force:

• \( F = \frac{kq_1q_2}{r^2} \)
• Where \( F \) is the force exerted between two point charges \( q_1 \) and \( q_2 \).
• \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).
• \( r \) is the distance separating the two charges.

If the charges are like (both positive or both negative), the force is repulsive, pushing them apart. If they are unlike (one positive and one negative), the force is attractive, pulling them together.

In electrostatics, this force plays a major role in determining the behavior of charges and the electric fields they create around them. Precise calculation of these forces can predict how charged objects will interact in various scenarios, such as in the presence of a uniformly charged rod as posed in the exercise.

Integration in Electrostatics

Integration is a powerful tool in electrostatics for determining the total effect of continuous charge distributions. Unlike discrete charges, a continuous charge distribution, such as a charged rod, requires the use of integration to sum up the infinitesimal contributions from each small charge element.

In this exercise, to find the total electric force exerted by a charged rod on a point charge, we set up an infinitesimal charge element \( dq \):

• \( dq = \lambda dx \)
• Where \( \lambda \) is the linear charge density, \( dx \) is a small segment along the rod.

The small electric force \( df \) exerted by \( dq \) on the point charge is given by:

• \( df = \frac{kq dq}{(r+x)^2} \)
• Where \( r+x \) is the distance from the point charge to the element \( dq \).

To find the total force \( F \), integrate \( df \) over the entire length of the rod:

• \( F = \int_0^L df = \frac{kqQ}{L} \int_0^L \frac{dx}{(r+x)^2} \)

The integral accounts for contributions from all infinitesimal elements, giving the net force on the point charge. This method is essential in physics to handle real-world systems with charge distributions that are not simply point charges.

Linear Charge Density

Linear charge density is an important concept in physics, as it describes how charge is distributed along a line or a linear object like a rod. It is defined as the charge per unit length and is denoted by the symbol \( \lambda \).

For a rod of total charge \( Q \) and length \( L \), the linear charge density \( \lambda \) is given by:

• \( \lambda = \frac{Q}{L} \)
• This expression indicates that the charge is uniformly distributed across the entire length of the rod.

Understanding and determining linear charge density is crucial when dealing with continuous charge distributions, as it allows us to calculate the effect of each small section of the charge on other objects.

When working with charge densities:

• \( dq = \lambda dx \) gives the infinitesimal amount of charge in a small segment \( dx \).
• This simplifies the calculation of electric effects like force or electric field by dividing the whole problem into manageable segments.

By breaking down continuous distributions into small units, linear charge density aids in effectively applying integration methods to solve electrostatic problems, as seen in this exercise with the charged rod and point charge interaction.