/*! 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 91 A van de Graaff generator has a ... [FREE SOLUTION] | 91影视

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Chapter 17: Problem 91

A van de Graaff generator has a metal sphere of radius \(15 \mathrm{cm} .\) To what potential can it be charged before the electric field at its surface exceeds \(3.0 \times 10^{6} \mathrm{N} / \mathrm{C}\) (which is sufficient to break down dry air and initiate a spark)?

Short Answer

Expert verified
Answer: The potential of the metal sphere would be approximately \(4.5 \times 10^4\, \mathrm{V}\) before the electric field at its surface exceeds \(3.0 \times 10^6\, \mathrm{N/C}\).

Step by step solution

01

Express the electric field as a function of charge and radius

To find the electric field on the surface of the sphere, we can use the following equation: $$ E = \frac{kQ}{R^{2}}, $$ where \(E\) is the electric field, \(k\) is the electrostatic constant (\(k \approx 8.99 \times 10^{9} \mathrm{N m^{2} C^{-2}}\)), \(Q\) is the charge on the sphere, and \(R\) is the radius of the sphere. We are given that \(E = 3.0 \times 10^{6} \mathrm{N/C}\) and \(R = 15 \mathrm{cm} = 0.15 \mathrm{m}\). Our goal is to find \(Q\).
02

Solve for the charge\(Q\) of the sphere

Replacing the electric field and the radius in the electric field equation and solving for \(Q\), we get: $$ Q = \frac{E R^{2}}{k}. $$ Now, plug in the values provided: $$ Q = \frac{(3.0 \times 10^{6} \mathrm{N/C})\times(0.15 \mathrm{m})^{2}}{8.99 \times 10^{9} \mathrm{N m^{2} C^{-2}}}\approx 7.5 \times 10^{-5}\,\mathrm{C}. $$ So, the charge on the sphere before the electric field exceeds \(3.0 \times 10^6 \mathrm{N/C}\) is approximately \(7.5 \times 10^{-5} \mathrm{C}\).
03

Determine the electric potential of the sphere

Given that we found the charge on the sphere, we can now calculate the electric potential (\(V\)) using the following formula: $$ V = \frac{kQ}{R}, $$ where \(V\) is the electric potential, \(k\) is the electrostatic constant, \(Q\) is the charge on the sphere, and \(R\) is the radius of the sphere.
04

Calculate the electric potential

Replacing the charge \(Q\) and the radius in the electric potential equation, we have: $$ V = \frac{(8.99 \times 10^{9} \mathrm{N m^{2} C^{-2}})\times(7.5 \times 10^{-5}\,\mathrm{C})}{0.15 \mathrm{m}}\approx 4.5 \times 10^4\,\mathrm{V}. $$ Hence, the sphere can be charged up to a potential of approximately \(4.5 \times 10^4\,\mathrm{V}\) before the electric field at its surface exceeds \(3.0 \times 10^6 \mathrm{N/C}\).

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Most popular questions from this chapter

A parallel plate capacitor is composed of two square plates, $10.0 \mathrm{cm}\( on a side, separated by an air gap of \)0.75 \mathrm{mm} .$ (a) What is the charge on this capacitor when there is a potential difference of \(150 \mathrm{V}\) between the plates? (b) What energy is stored in this capacitor?
A cell membrane has a surface area of \(1.0 \times 10^{-7} \mathrm{m}^{2},\) a dielectric constant of \(5.2,\) and a thickness of \(7.5 \mathrm{nm}\) The membrane acts like the dielectric in a parallel plate capacitor; a layer of positive ions on the outer surface and a layer of negative ions on the inner surface act as the capacitor plates. The potential difference between the "plates" is \(90.0 \mathrm{mV}\). (a) How much energy is stored in this capacitor? (b) How many positive ions are there on the outside of the membrane? Assume that all the ions are singly charged (charge +e).
A shark is able to detect the presence of electric fields as small as $1.0 \mu \mathrm{V} / \mathrm{m} .$ To get an idea of the magnitude of this field, suppose you have a parallel plate capacitor connected to a 1.5 - \(V\) battery. How far apart must the parallel plates be to have an electric field of $1.0 \mu \mathrm{V} / \mathrm{m}$ between the plates?
A parallel plate capacitor is attached to a battery that supplies a constant voltage. While the battery is still attached, a dielectric of dielectric constant \(\kappa=3.0\) is inserted so that it just fits between the plates. What is the energy stored in the capacitor after the dielectric is inserted in terms of the energy \(U_{0}\) before the dielectric was inserted?

A parallel plate capacitor is attached to a battery that supplies a constant voltage. While the battery remains attached to the capacitor, the distance between the parallel plates increases by \(25 \% .\) What happens to the energy stored in the capacitor?
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