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Chapter 11: Problem 1522

A copper sphere of mass \(2 \mathrm{gm}\) contains about \(2 \times 10^{22}\) atoms. The charge on the nucleus of each atom is 29e. what fraction of electrons removed from the sphere to give it a charge of \(2 \mu \mathrm{c}\) ? (A) \(2 \times 10^{-10}\) (B) \(1.19 \times 10^{-12}\) (C) \(1.25 \times 10^{-11}\) (D) \(2.16 \times 10^{-11}\)

Short Answer

Expert verified
The short answer is (B) \(1.19 \times 10^{-12}\).

Step by step solution

01

Calculate the total number of electrons in the sphere

The copper sphere contains about \(2 \times 10^{22}\) atoms, and the charge on the nucleus of each atom is \(29e\), where \(e\) is the elementary charge, \(1.6 \times 10^{-19} C\). Therefore, each copper atom has 29 electrons. To find the total number of electrons in the sphere, we need to multiply the number of atoms by the number of electrons per atom: \[ \text{Total electrons} = (\text{Number of atoms})(\text{Number of electrons per atom}) = (2 \times 10^{22})(29) \]
02

Calculate the number of electrons to be removed

Given that we need to give the sphere a charge of \(2\mu C\), we first need to find out how many electrons account for this charge. We can calculate this by dividing the desired charge by the charge of a single electron: \[ \text{Number of electrons to be removed} = \frac{\text{Desired charge}}{e} = \frac{2\mu C}{1.6 \times 10^{-19} C} = \frac{2 \times 10^{-6} C}{1.6 \times 10^{-19} C} \]
03

Calculate the fraction of electrons removed

Now that we have the total number of electrons and the number of electrons to be removed, we can find the fraction of electrons that need to be removed by dividing the number of electrons to be removed by the total number of electrons: \[ \text{Fraction of electrons removed} = \frac{\text{Number of electrons to be removed}}{\text{Total electrons}} = \frac{\frac{2 \times 10^{-6} C}{1.6 \times 10^{-19} C}}{(2 \times 10^{22})(29)} \]
04

Simplify and find the correct answer

Simplify the fraction and compare it with the provided options: \[ \text{Fraction of electrons removed} = \frac{2 \times 10^{-6}}{1.6 \times 10^{-19} \times 2 \times 10^{22} \times 29} = 1.19 \times 10^{-12} \] So, the answer is: (B) \(1.19 \times 10^{-12}\)

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Most popular questions from this chapter

Three charges \(2 q,-q,-q\) are located at the vertices of an equilateral triangle. At the centre of the triangle. (A) The Field is Zero but Potential is non - zero (B) The Field is non - Zero but Potential is zero (C) Both field and Potential are Zero (D) Both field and Potential are non - Zero

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 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.

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) \(i \wedge\left(2 x y-z^{3}\right)+j \wedge x y^{2}+k \wedge 3 z^{2} x\)

Iaking earth to be a metallic spheres, its capacity will approximately be (A) \(6.4 \times 10^{6} \mathrm{~F}\) (B) \(700 \mathrm{pF}\) (C) \(711 \mu \mathrm{F}\) (D) \(700 \mathrm{pF}\)
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