/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 69 A body has a charge of one coulo... [FREE SOLUTION] | 91影视

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Chapter 13: Problem 69

A body has a charge of one coulomb. The number of excess (or lesser) electrons on it from its normal state will be (A) \(\infty\) (B) \(1.6 \times 10^{-19}\) (C) \(1.6 \times 10^{19}\) (D) \(6.25 \times 10^{18}\)

Short Answer

Expert verified
The number of excess (or lesser) electrons on the body from its normal state is \(6.25 脳 10^{18}\). So, the correct answer is (D) \(6.25 \times 10^{18}\).

Step by step solution

01

Identifying the given information

We know the following information: - Charge of the body, Q = 1 C (coulomb) - Charge of one electron, e = \(-1.6 \times 10^{-19}\) C - We need to find the number of excess or lesser electrons, n Note that the charge of an electron is negative, indicating that the body has an excess of electrons.
02

Use the formula of charge

The formula to calculate the charge of a body with a certain number of excess or lesser electrons is: Q = n 脳 e Where, Q = charge of the body n = number of excess or lesser electrons e = charge of one electron Our given information is: Q = 1 C e = \(-1.6 \times 10^{-19}\) C We need to find n.
03

Solve for n

We can now solve the equation Q = n 脳 e for n: 1 C = n 脳 \((-1.6 \times 10^{-19})\) C To find n, divide both sides by \(-1.6 \times 10^{-19}\) C: n = \(\frac{1 C}{-1.6 \times 10^{-19} C}\)
04

Calculate the number of excess or lesser electrons

We can now calculate the value of n: n = \(\frac{1}{-1.6 \times 10^{-19}}\) n = \(6.25 脳 10^{18}\) The number of excess (or lesser) electrons on the body from its normal state is \(6.25 脳 10^{18}\). So, the correct answer is (D) \(6.25 \times 10^{18}\).

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Most popular questions from this chapter

A charge \(Q\) is placed at the opposite corners of a square. A charge \(q\) is placed at each of the other two corners. If the net electrical force on \(Q\) is zero, then the \(\frac{Q}{q}\) equals (A) \(-2 \sqrt{2}\) (B) \(-1\) (C) 1 (D) \(-\frac{1}{\sqrt{2}}\)

Suppose the electrostatic potential at some points in space are given by \(V=\left(x^{2}-2 x\right)\). The electrostatic field strength at \(x=1\) is (A) Zero (B) \(-2\) (C) 2 (D) 4

Four point charges \(q_{1}, q_{2}, q_{3}\), and \(q_{4}\) are placed at the corners of the square of side \(\mathrm{a}\), as shown in Fig. 13.36. The potential at the centre of the square is (Given: \(q_{1}=1 \times 10^{-8} \mathrm{C}, q_{2}=-2 \times 10^{-8} \mathrm{C}\), \(\left.q_{3}=3 \times 10^{-8} \mathrm{C}, q_{4}=2 \times 10^{-8} \mathrm{C}, a=1 \mathrm{~m}\right)\) (A) \(507 \mathrm{~V}\) (B) \(607 \mathrm{~V}\) (C) \(550 \mathrm{~V}\) (D) \(650 \mathrm{~V}\)

Two plates are \(2 \mathrm{~cm}\) apart. A potential difference of \(10 \mathrm{~V}\) is applied between them, the electric field between the plates is (A) \(20 \mathrm{~N} / \mathrm{C}\) (B) \(500 \mathrm{~N} / \mathrm{C}\) (C) \(5 \mathrm{~N} / \mathrm{C}\) (D) \(250 \mathrm{~N} / \mathrm{C}\)

A charged sphere of diameter \(4 \mathrm{~cm}\) has a charge density of \(10^{-4}\) coulombs/cm \(^{2}\). The work done in joules when a charge of 40 nano-coulombs is moved from infinity to a point which is at a distance of \(2 \mathrm{~cm}\) from the surface of the sphere is (A) \(14.4 \pi\) (B) \(28.8 \pi\) (C) \(144 \pi\) (D) \(288 \pi\)
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