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Chapter 11: Problem 1563

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) \(-4 \times 10^{3} \mathrm{E}_{0}\)

Short Answer

Expert verified
The total charge inside the surface is approximately \(\left[-4 \times 10^{3}/\epsilon_{0}\right]\), which is closest to option (A). However, the exact value of the total charge enclosed is \(-3.54 \times 10^{-8} \ \mathrm{C}\).

Step by step solution

01

Given information

We are given the inward and outward electric fluxes for a closed surface. The inward electric flux is \(8 \times 10^3 \ \mathrm{Nm}^2/\mathrm{C}\), and the outward electric flux is \(4 \times 10^3 \ \mathrm{Nm}^2/\mathrm{C}\).
02

Applying Gauss's Law

Gauss's Law states that the total electric flux through a closed surface is equal to the total charge enclosed by the surface divided by the permittivity of free space (\(\epsilon_0\)). Mathematically, it can be written as: \[\Phi_E = \frac{Q_{enclosed}}{\epsilon_0}\] Where \(\Phi_E\) is the total electric flux through the closed surface, \(Q_{enclosed}\) is the total charge enclosed by the surface, and \(\epsilon_0\) is the permittivity of free space (\(\epsilon_0 \approx 8.85 \times 10^{-12} \ \mathrm{C}^2/\mathrm{Nm}^2\)).
03

Find the net electric flux

We need to find the net electric flux through the closed surface. Since the inward flux is negative and the outward flux is positive, we subtract the inward flux from the outward flux: \[\Phi_{net} = \Phi_{outward} - \Phi_{inward}\] \[\Phi_{net} = 4\times 10^3 - 8\times 10^3 = - 4\times 10^3 \ \mathrm{Nm}^2/\mathrm{C}\]
04

Calculate the total charge enclosed

Now, we can use Gauss's Law to find the total charge enclosed by the surface: \[Q_{enclosed} = \Phi_{net} \cdot \epsilon_0\] \[Q_{enclosed} = (-4 \times 10^3 \ \mathrm{Nm}^2/\mathrm{C}) \cdot (8.85 \times 10^{-12} \ \mathrm{C}^2/\mathrm{Nm}^2)\] \[Q_{enclosed} = -3.54 \times 10^{-8} \ \mathrm{C}\] We can see that the total charge enclosed by the surface is closest to option (A): (A) \(\left[-4 \times 10^{3}/\epsilon_{0}\right]\) However, the \(Q_{enclosed}\approx -3.54 \times 10^{-8}\), so the correct answer is closest to, but not exactly equal to option (A).

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

A simple pendulum consists of a small sphere of mass \(\mathrm{m}\) suspended by a thread of length \(\ell\). The sphere carries a positive charge q. The pendulum is placed in a uniform electric field of strength \(\mathrm{E}\) directed Vertically upwards. If the electrostatic force acting on the sphere is less than gravitational force the period of pendulum is (A) $\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}-(\mathrm{q} \mathrm{E} / \mathrm{m})\\}]^{(1 / 2)}$ (B) \(\mathrm{T}=2 \pi(\ell / \mathrm{g})^{(1 / 2)}\) \(\left.\left.\left.\mathrm{m}_{\mathrm{}}\right\\}\right\\}\right]^{(1 / 2)}\) (D) \(\mathrm{T}=2 \pi[(\mathrm{m} \ell / \mathrm{qE})]^{(1 / 2)}\) (C) \(\mathrm{T}=2 \pi[\ell /\\{\mathrm{g}+(\mathrm{qE} / \mathrm{t}\)

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The electric Potential at a point $\mathrm{P}(\mathrm{x}, \mathrm{y}, \mathrm{z})\( is given by \)\mathrm{V}=-\mathrm{x}^{2} \mathrm{y}-\mathrm{x} \mathrm{z}^{3}+4\(. The electric field \)\mathrm{E}^{\boldsymbol{T}}$ at that point is \(\ldots \ldots\) (A) $i \wedge\left(2 \mathrm{xy}+\mathrm{z}^{3}\right)+\mathrm{j} \wedge \mathrm{x}^{2}+\mathrm{k} \wedge 3 \mathrm{xz}^{2}$ (B) $i \wedge 2 \mathrm{xy}+\mathrm{j} \wedge\left(\mathrm{x}^{2}+\mathrm{y}^{2}\right)+\mathrm{k} \wedge\left(3 \mathrm{xy}-\mathrm{y}^{2}\right)$ (C) \(i \wedge z^{3}+j \wedge x y z+k \wedge z^{2}\) (D) \(i \wedge\left(2 x y-z^{3}\right)+j \wedge x y^{2}+k \wedge 3 z^{2} x\)

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