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Chapter 17: Problem 12

The electric strength of air is \(2 \times 10^{7} \mathrm{NC}^{-1}\). The maximum charge that a metaleic sphere of diameter \(6 \mathrm{~mm}\) can hold is (a) \(3 \mathrm{nC}\) (b) \(20 \mathrm{nC}\) (c) \(1.5 \mathrm{nC}\) (d) \(2 \mathrm{nC}\)

Short Answer

Expert verified
The maximum charge the sphere can hold is 2 nC, so option (d) is correct.

Step by step solution

01

Understand the Problem

We need to find the maximum charge a metal sphere can hold before air breaks down and loses its insulating properties. The electric field on the surface of a sphere is connected to the charge and radius of the sphere.
02

Identify Given Values

We know the electric strength of air, which is the maximum electric field air can sustain, is given as \(2 \times 10^{7} \mathrm{NC}^{-1}\). The diameter of the sphere is \(6 \mathrm{~mm}\), so the radius \(r\) is \(3 \mathrm{~mm}\) or \(3 \times 10^{-3} \mathrm{~m}\).
03

Relate Electric Field and Charge

The electric field \(E\) on the surface of a sphere is given by \(E = \frac{kQ}{r^2}\), where \(k\) is the Coulomb's constant \(8.99 \times 10^9 \mathrm{Nm}^2/ ext{C}^2\), \(Q\) is the charge, and \(r\) is the radius.
04

Set the Electric Field Equal to Air Strength

To find the maximum charge before breakdown, set \(E = 2 \times 10^{7} \mathrm{NC}^{-1}\). Thus, \(\frac{kQ}{r^2} = 2 \times 10^{7}\).
05

Solve For Charge \(Q\)

Rearrange the equation \(Q = \frac{E \cdot r^2}{k}\). Substitute the values to compute \(Q\): \[Q = \frac{2 \times 10^{7} \cdot (3 \times 10^{-3})^2}{8.99 \times 10^9}\] \[Q = \frac{2 \times 10^{7} \cdot 9 \times 10^{-6}}{8.99 \times 10^9}\] \[Q = \frac{1.8 \times 10^2}{8.99 \times 10^9}\] \[Q \approx 2 \times 10^{-9} \mathrm{C} = 2 \mathrm{nC}\]
06

Choose the Correct Option

The calculated maximum charge \(Q\) is \(2 \mathrm{nC}\). Therefore, the correct answer is option (d) \(2 \mathrm{nC}\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

electric strength of air
The electric strength of air is a crucial concept in understanding how charges behave in our environment. It represents the maximum electric field that air can withstand before it starts conducting electricity.
When this threshold is exceeded, air loses its role as an insulator and becomes a conductor.
  • The electric strength of air is typically around \(2 \times 10^{7} \mathrm{NC}^{-1}\).
  • This value indicates the point at which ionization can occur, leading to phenomena such as sparks or lightning.
Understanding this helps in applications such as designing high-voltage equipment and ensuring electrical safety.
For instance, it is why overhead wires are placed at safe distances apart to prevent unintended discharge of electricity through the air.
charge on metal sphere
The charge on a metal sphere is influenced by several factors, including its size and the surrounding electric field.
When we talk about finding the charge a sphere can hold, it's about determining the limit before unwanted electric discharge occurs.
  • The geometry of a sphere means that charge uniformly distributes over its surface.
  • The radius of the sphere directly impacts the electric field on its surface.
For a given electric field strength, a larger sphere can generally hold more charge than a smaller one, because the total surface where the charge distributes is greater.
Therefore, understanding the sphere's properties and how they relate to electric fields can help calculate how much charge it can sustain safely.
Coulomb's law
Coulomb's Law provides the mathematical framework to understand the electric forces between charges.
It states that the electric force \(F\) between two point charges, \(Q_1\) and \(Q_2\), separated by a distance \(r\), is proportional to the product of their charges and inversely proportional to the square of the distance between them.
\[ F = k \frac{Q_1 Q_2}{r^2} \] Here, \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \mathrm{Nm}^2/\mathrm{C}^2\).
  • This law is fundamental in electrostatics, describing how charged objects interact in space.
  • It demonstrates that as the distance between charges increases, the force decreases significantly.
The application of Coulomb's law is essential in solving problems related to electric fields, as it was in determining the maximum charge a sphere can hold before breakdown occurs, providing a practical insight into electrostatic phenomena.

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Most popular questions from this chapter

A charge \((-q)\) and another charge \((+Q)\) are kept a two points \(A\) and \(B\), respectively. Keeping the charg \((+Q)\) fixed at \(B\), the charge \((-q)\) at \(A\) is moved \(t\) another point \(C\) such that \(A B C\) forms an equilatera triangle of side \(l\). The net work done in moving th charge \((-q)\) is (a) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q}{l}\) (b) \(\frac{1}{4 \pi \varepsilon_{0}} \frac{Q q}{l^{2}}\) (c) \(\frac{1}{4 \pi \varepsilon_{0}} Q q l\) (d) zero

A charged body has an electric flux \(\phi\) associated with it. The body is now placed inside a metallic container. The electric flux, \(\phi_{1}\) associated with the container will be (a) \(\phi_{1}=0\) (b) \(0<\phi_{1}<\phi\) (c) \(\phi_{1}=\phi\) (d) \(\phi_{1} \geq \phi\)

The electrostatic potential inside a charged spherical ball is given by \(\phi=a r^{2}+b\), where, \(r\) is the distance from the centre, \(a\) and \(b\) are constants. Then the charge density inside the ball is (a) \(-24 \pi a \varepsilon_{0} r\) (b) \(-6 a \varepsilon_{0}\) (c) \(-24 \pi \varepsilon_{0}\) (d) \(-6 a \varepsilon_{0} r\)

An electron of mass \(M_{e}\), initially at rest, moves through a certain distance in a uniform electric field in time \(t_{1}\). A proton of mass \(M_{p}\) also initially at rest, takes time \(t_{2}\) to move through an equal distance in this uniform electric field. Neglecting the effect of gravity, the ratio \(t_{2} / t_{1}\) is nearly equal to (a) 1 (b) \(\sqrt{M_{p} / M_{e}}\) (c) \(\sqrt{M_{e} / M_{p}}\) (d) 1836

The charge of \(+\frac{10}{3} \times 10^{-9} \mathrm{C}\) are placed at each of the four corners of a square of side \(8 \mathrm{~cm}\). The potential at the point of intersection of the diagonals, is (a) \(1500 \sqrt{2} \mathrm{~V}\) (b) \(1800 \sqrt{2} \mathrm{~V}(\mathrm{c}) 600 \sqrt{2} \mathrm{~V}\) (d) \(900 \sqrt{2} \mathrm{~V}\)
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