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Chapter 11: Problem 1539

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

Expert verified
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

01

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 \(\lambda=\frac{Q}{L}\).
02

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: \(df = \frac{kq(Q/L)dx}{(r+x)^2}\)
03

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: \(F = \frac{kqQ}{L} \int_r^{r+L} \frac{du}{u^2} = -\frac{kqQ}{L} \left[ \frac{1}{u} \right]_r^{r+L}\)
04

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: \(\boxed{(B) \frac{kqQ}{r(r+L)}}\)

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Most popular questions from this chapter

A ball of mass \(1 \mathrm{gm}\) and charge \(10^{-8} \mathrm{c}\) moves from a point \(\mathrm{A}\), where the potential is 600 volt to the point \(B\) where the potential is zero. Velocity of the ball of the point \(\mathrm{B}\) is $20 \mathrm{~cm} / \mathrm{s}\(. The velocity of the ball at the point \)\mathrm{A}$ will be \(\ldots \ldots\) (A) \(16.8(\mathrm{~m} / \mathrm{s})\) (B) \(22.8(\mathrm{~cm} / \mathrm{s})\) (C) \(228(\mathrm{~cm} / \mathrm{s})\) (D) \(168(\mathrm{~m} / \mathrm{s})\)

N identical drops of mercury are charged simultaneously to 10 volt. when combined to form one large drop, the potential is found to be 40 volt, the value of \(\mathrm{N}\) is \(\ldots \ldots\) (A) 4 (B) 6 (C) 8 (D) 10

Two charged spheres of radii \(R_{1}\) and \(R_{2}\) having equal surface charge density. The ratio of their potential is ..... (A) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)\) (B) \(\left(\mathrm{R}_{2} / \mathrm{R}_{1}\right)^{2}\) (C) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)^{2}\) (D) \(\left(\mathrm{R}_{1} / \mathrm{R}_{2}\right)\)

The circular plates \(\mathrm{A}\) and \(\mathrm{B}\) of a parallel plate air capacitor have a diameter of \(0.1 \mathrm{~m}\) and are $2 \times 10^{-3} \mathrm{~m}\( apart. The plates \)\mathrm{C}\( and \)\mathrm{D}$ of a similar capacitor have a diameter of \(0.1 \mathrm{~m}\) and are $3 \times 10^{-3} \mathrm{~m}\( apart. Plate \)\mathrm{A}\( is earthed. Plates \)\mathrm{B}$ and \(\mathrm{D}\) are connected together. Plate \(\mathrm{C}\) is connected to the positive pole of a \(120 \mathrm{~V}\) battery whose negative is earthed, The energy stored in the system is (A) \(0.1224 \mu \mathrm{J}\) (B) \(0.2224 \mu \mathrm{J}\) (C) \(0.4224 \mu \mathrm{J}\) (D) \(0.3224 \mu \mathrm{J}\)

A simple pendulum of period \(\mathrm{T}\) has a metal bob which is negatively charged. If it is allowed to oscillate above a positively charged metal plate, its period will ...... (A) Remains equal to \(\mathrm{T}\) (B) Less than \(\mathrm{T}\) (C) Infinite (D) Greater than \(\mathrm{T}\)
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