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Chapter 11: Problem 1608

Charges \(+q\) and \(-q\) are placed at point \(A\) and \(B\) respectively which are a distance \(2 \mathrm{~L}\) apart, \(\mathrm{C}\) is the midpoint between \(\mathrm{A}\) and \(\mathrm{B}\). The work done in moving a charge \(+Q\) along the semicircle \(C R D\) is \(\ldots \ldots\) (A) $\left[(\mathrm{qQ}) /\left(2 \pi \mathrm{\epsilon}_{0} \mathrm{~L}\right)\right]$ (B) \(\left[(-q Q) /\left(6 \pi \in_{0} L\right)\right]\) (C) $\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]$ (D) \(\left[(q Q) /\left(4 \pi \in_{0} L\right)\right]\)

Short Answer

Expert verified
The work done in moving a charge \(+Q\) along the semicircle \(CRD\) is approximately \(\left[(\mathrm{qQ}) /\left(6 \pi \mathrm{e}_{0} \mathrm{~L}\right)\right]\), although none of the given answer choices precisely equal to zero. However, option (C) is the most similar to the correct value.

Step by step solution

01

1. Identify the given information

: We are given: - Charge \(+q\) at point A - Charge \(-q\) at point B - The distance between point A and B is \(2L\) - Point C is the midpoint between point A and point B - Charge \(+Q\) is moved along the semicircle \(CRD\)
02

2. Determine the electric field produced by the charges

: The electric field produced by a single charge is given by: \[E = \frac{k\cdot q}{r^2}\] where - \(E\) is the electric field - \(k\) is the electric constant, \(k = \frac{1}{4\pi\epsilon_0}\) - \(q\) is the charge - \(r\) is the distance between the charge and the point of interest
03

3. Consider the symmetry of the situation

: Since point C is the midpoint between point A and point B, the electric field from charge \(+q\) at point A and charge \(-q\) at point B are equal and opposite at point C. Hence, the electric fields at point C cancel each other and the net field is zero. As the charge \(+Q\) moves along the semicircle, the same cancellation occurs due to symmetry.
04

4. Calculate the work done

: Since the electric field at all points on the semicircle is zero due to the symmetry of the situation, the work done on charge \(+Q\) in moving along the semicircle \(CRD\) is also zero. Therefore, the work done is: \[W = 0\] Considering the given options, none of them is precisely equal to zero. However, among the options, (A) and (C) are positive, while (B) and (D) are negative. It can be interpreted that the options are missing a sign and thus, option (C) is the most similar to zero, so we can consider it to be the correct answer when compared to the other given choices.

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Most popular questions from this chapter

Two point masses \(\mathrm{m}\) each carrying charge \(-\mathrm{q}\) and \(+\mathrm{q}\) are attached to the ends of a massless rigid non-conducting rod of length \(\ell\). The arrangement is placed in a uniform electric field \(\mathrm{E}\) such that the rod makes a small angle \(5^{\circ}\) with the field direction. The minimum time needed by the rod to align itself along the field is \(\ldots \ldots .\) (A) \(t=\pi \sqrt{[}(2 M \ell) /(3 q E)]\) (B) $\mathrm{t}=(\pi / 2) \sqrt{[}(\mathrm{M} \ell) /(2 \mathrm{q} \mathrm{E})]$ (D) \(t=2 \pi \sqrt{[}(M \ell / E)]\)

A parallel plate condenser with dielectric of constant \(\mathrm{K}\) between the plates has a capacity \(\mathrm{C}\) and is charged to potential \(\mathrm{v}\) volt. The dielectric slab is slowly removed from between the plates and reinserted. The net work done by the system in this process is (A) Zero (B) \((1 / 2)(\mathrm{K}-1) \mathrm{cv}^{2}\) (C) \((\mathrm{K}-1) \mathrm{cv}^{2}\) (D) \(\mathrm{cv}^{2}[(\mathrm{~K}-1) / \mathrm{k}]\)

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) $\left[\left(\mathrm{A} \in{ }_{0} \mathrm{~V}_{0}{ }^{2}\right) /(4 \mathrm{~d})\right]$

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) \(168(\mathrm{~m} / \mathrm{s})\)

Charges of \(+(10 / 3) \times 10^{-9} \mathrm{C}\) are placed at each of the four corners of a square of side \(8 \mathrm{~cm}\). The potential at the intersection of the diagonals is ...... (A) \(150 \sqrt{2}\) Volt (B) \(900 \sqrt{2}\) Volt (C) \(1500 \sqrt{2}\) Volt (D) \(900 \sqrt{2} \cdot \sqrt{2}\) Volt
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