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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

Two point charges of \(+16 \mathrm{c}\) and \(-9 \mathrm{c}\) are placed $8 \mathrm{~cm}\( apart in air \)\ldots \ldots\(.. distance of a point from \)-9$ c charge at which the resultant electric field is zero. (A) \(24 \mathrm{~cm}\) (B) \(9 \mathrm{~cm}\) (C) \(16 \mathrm{~cm}\) (D) \(35 \mathrm{~cm}\)

Two identical charged spheres suspended from a common point by two massless strings of length \(\ell\) are initially a distance d \((d<<\ell)\) apart because of their mutual repulsion. The charge begins to leak from both the spheres at a constant rate. As a result the spheres approach each other with a velocity \(\mathrm{v}\). Then function of distance \(\mathrm{x}\) between them becomes \(\ldots \ldots\) (A) \(v \propto x\) (B) \(\mathrm{v} \propto \mathrm{x}^{(-1 / 2)}\) (C) \(\mathrm{v} \propto \mathrm{x}^{-1}\) (D) \(\mathrm{v} \propto \mathrm{x}^{(1 / 2)}\)

Two point charges \(-q\) and \(+q\) are located at points \((0,0,-a)\) and $(0,0, a)\( respectively. The potential at a point \)(0,0, z)\( where \)z>a\( is \)\ldots \ldots$ (A) $\left[(2 \mathrm{q} a) /\left\\{4 \pi \epsilon_{0}\left(z^{2}+a^{2}\right)\right\\}\right]$ (B) \(\left[\mathrm{q} /\left(4 \pi \epsilon_{0} \mathrm{a}\right)\right]\) (C) \(\left[\right.\) (qa) \(\left./\left(4 \pi \in_{0} z^{2}\right)\right]\) (D) $\left[(2 q a) /\left\\{4 \pi \epsilon_{0}\left(z^{2}-a^{2}\right)\right\\}\right]$

Two parallel plate air capacitors have their plate areas 100 and $500 \mathrm{~cm}^{2}$ respectively. If they have the same charge and potential and the distance between the plates of the first capacitor is \(0.5 \mathrm{~mm}\), what is the distance between the plates of the second capacitor ? (A) \(0.25 \mathrm{~cm}\) (B) \(0.50 \mathrm{~cm}\) (C) \(0.75 \mathrm{~cm}\) (D) \(1 \mathrm{~cm}\)

The inward and outward electric flux for a closed surface in units of \(\mathrm{Nm}^{2} / \mathrm{C}\) are respectively \(8 \times 10^{3}\) and $4 \times 10^{3}\(. Then the total charge inside the surface is \)\ldots \ldots \ldots \ldots . . \mathrm{c}$. (A) \(\left[\left(-4 \times 10^{3}\right) / \epsilon_{0}\right]\) (B) \(-4 \times 10^{3}\) (C) \(4 \times 10^{3}\) (D) \(-4 \times 10^{3} \mathrm{E}_{0}\)
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