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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

Before a lightning strike can occur, the breakdown limit for damp air must be reached. If this occurs for an electric field of $3.33 \times 10^{5} \mathrm{V} / \mathrm{m},$ what is the maximum possible height above the Earth for the bottom of a thundercloud, which is at a potential $1.00 \times 10^{8} \mathrm{V}$ below Earth's surface potential, if there is to be a lightning strike?
A parallel plate capacitor has a charge of \(0.020 \mu \mathrm{C}\) on each plate with a potential difference of \(240 \mathrm{V}\). The parallel plates are separated by 0.40 mm of air. What energy is stored in this capacitor?
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 certain capacitor stores \(450 \mathrm{J}\) of energy when it holds $8.0 \times 10^{-2} \mathrm{C}$ of charge. What is (a) the capacitance of this capacitor and (b) the potential difference across the plates?
Two metal spheres are separated by a distance of \(1.0 \mathrm{cm}\) and a power supply maintains a constant potential difference of \(900 \mathrm{V}\) between them. The spheres are brought closer to one another until a spark flies between them. If the dielectric strength of dry air is $3.0 \times 10^{6} \mathrm{V} / \mathrm{m},$ what is the distance between the spheres at this time?
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