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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

Two identical balls having like charges and placed at a certain distance apart repel each other with a certain force. They are brought in contact and then moved apart to a distance equal to half their initial separation. The force of repulsion between them increases \(4.5\) times in comparison with the initial value. The ratio of the initial charges of the balls is ....... (A) \(4: 1\) (B) \(6: 1\) (C) \(3: 1\) (D) \(2: 1\)

Large number of capacitors of rating $10 \mu \mathrm{F} / 200 \mathrm{~V} \mathrm{~V}$ are available. The minimum number of capacitors required to design a \(10 \mu \mathrm{F} / 700 \mathrm{~V}\) capacitor is (A) 16 (B) 8 (C) 4 (D) 7

A thin spherical shell of radius \(R\) has charge \(Q\) spread uniformly over its surface. Which of the following graphs, figure most closely represents the electric field \(\mathrm{E}\) (r) produced by the shell in the range $0 \leq \mathrm{r}<\infty\(, where \)\mathrm{r}$ is the distance from the centre of the shel1.

A charged particle of mass \(\mathrm{m}\) and charge \(q\) is released from rest in a uniform electric field \(E\). Neglecting the effect of gravity, the kinetic energy of the charged particle after 't' second is ...... (A) $\left[\left(\mathrm{Eq}^{2} \mathrm{~m}\right) /\left(2 \mathrm{t}^{2}\right)\right]$ (B) $\left[\left(\mathrm{E}^{2} \mathrm{q}^{2} \mathrm{t}^{2}\right) /(2 \mathrm{~m})\right]$ (C) \(\left[\left(2 \mathrm{E}^{2} \mathrm{t}^{2}\right) /(\mathrm{qm})\right]\) (D) \([(\mathrm{Eqm}) / \mathrm{t}]\)

A parallel plate capacitor has plate of area \(\mathrm{A}\) and separation d. It is charged to a potential difference \(\mathrm{V}_{0}\). The charging battery is disconnected and the plates are pulled apart to three times the initial separation. The work required to separate the plates is (A) $\left[\left(\mathrm{A} \in_{0} \mathrm{~V}_{0}^{2}\right) / \mathrm{d}\right]$ (B) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}^{2}\right) /(2 \mathrm{~d})\right]$ (C) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(3 \mathrm{~d})\right]$ (D) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(4 \mathrm{~d})\right]$
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