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Chapter 11: Problem 1565

The electric flux for gaussian surface \(\mathrm{A}\) that enclose the $\ldots \ldots\( charged particles in free space is (given \)\left.q_{1}=-14 n c, q_{2}=78.85 \mathrm{nc}, q_{3}=-56 n c\right)$ (A) \(10^{4} \mathrm{Nm}^{2} / \mathrm{C}\) (B) \(10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (C) \(6.2 \times 10^{3} \mathrm{Nm}^{2} / \mathrm{C}\) (D) \(6.3 \times 10^{4} \mathrm{Nm}^{2} / \mathrm{C}\)

Short Answer

Expert verified
The electric flux for Gaussian surface A that encloses the charged particles in free space is \(10^3 \frac{N \cdot m^2}{C}\).

Step by step solution

01

Write down the formula for electric flux for enclosed charges

The Electric Flux formula through a closed surface is given by: \(\Phi = \frac{Q_{enclosed}}{\varepsilon_0}\) Where \(\Phi\) is the electric flux, \(Q_{enclosed}\) is the charge enclosed by the Gaussian surface and \(\varepsilon_0\) is the vacuum permittivity constant, approximately equal to \(8.85 \times 10^{-12}\, \frac{C^2}{N \cdot m^2}\).
02

Find the total enclosed charge

The enclosed charge is the sum of the given charges, \(q_1, q_2\), and \(q_3\). Calculate the total enclosed charge as follows: \(Q_{enclosed} = q_1 + q_2 + q_3\) Given the charges: \(q_1=-14\, nC\) \(q_2=78.85\, nC\) \(q_3=-56\, nC\)
03

Calculate the total enclosed charge

Replace the values of \(q_1, q_2\), and \(q_3\) in the equation and solve for \(Q_{enclosed}\): \(Q_{enclosed} = (-14 + 78.85 - 56) nC\) \(Q_{enclosed} = 8.85 nC\) Notice that the result is in nano Coulombs. We need to convert this value to Coulombs. \(1\,nC = 10^{-9}C\)
04

Convert the total enclosed charge to Coulombs

Convert the total enclosed charge to Coulombs: \(Q_{enclosed} = 8.85 \times 10^{-9} C\)
05

Calculate the electric flux

Now that we have the total enclosed charge in Coulombs, substitute the value of \(Q_{enclosed}\) into the electric flux formula and calculate \(\Phi\): \(\Phi = \frac{8.85 \times 10^{-9} C}{8.85 \times 10^{-12} \frac{C^2}{N \cdot m^2}}\) \(\Phi = 10^3 \frac{N \cdot m^2}{C}\) Comparing the calculated electric flux with the given options, we find that it matches option (B). The electric flux for Gaussian surface A that encloses the charged particles in free space is \(10^3 \frac{N \cdot m^2}{C}\).

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Most popular questions from this chapter

Electric potential at any point is $\mathrm{V}=-5 \mathrm{x}+3 \mathrm{y}+\sqrt{(15 \mathrm{z})}$, then the magnitude of the electric field is \(\ldots \ldots \ldots \mathrm{N} / \mathrm{C}\). (A) \(3 \sqrt{2}\) (B) \(4 \sqrt{2}\) (C) 7 (D) \(5 \sqrt{2}\)

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) \(\mathrm{cv}^{2}[(\mathrm{~K}-1) / \mathrm{k}]\)

Two point positive charges \(q\) each are placed at \((-a, 0)\) and \((a, 0)\). A third positive charge \(q_{0}\) is placed at \((0, y)\). For which value of \(\mathrm{y}\) the force at \(q_{0}\) is maximum \(\ldots \ldots \ldots\) (A) a (B) \(2 \mathrm{a}\) (C) \((\mathrm{a} / \sqrt{2})\) (D) \((\mathrm{a} / \sqrt{3})\)

Two metal plate form a parallel plate capacitor. The distance between the plates is \(\mathrm{d}\). A metal sheet of thickness \(\mathrm{d} / 2\) and of the same area is introduced between the plates. What is the ratio of the capacitance in the two cases? (A) \(4: 1\) (B) \(3: 1\) (C) \(2: 1\) (D) \(5: 1\)

The capacitors of capacitance \(4 \mu \mathrm{F}, 6 \mu \mathrm{F}\) and $12 \mu \mathrm{F}$ are connected first in series and then in parallel. What is the ratio of equivalent capacitance in the two cases? (A) \(2: 3\) (B) \(11: 1\) (C) \(1: 11\) (D) \(1: 3\)
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