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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 connected to a \(12-\mathrm{V}\) battery. While the battery remains connected, the plates are pushed together so the spacing is decreased. What is the effect on (a) the potential difference between the plates? (b) the electric field between the plates? (c) the magnitude of charge on the plates?
The potential difference across a cell membrane is \(-90 \mathrm{mV} .\) If the membrane's thickness is \(10 \mathrm{nm},\) what is the magnitude of the electric field in the membrane? Assume the field is uniform.
As an electron moves through a region of space, its speed decreases from $8.50 \times 10^{6} \mathrm{m} / \mathrm{s}\( to \)2.50 \times 10^{6} \mathrm{m} / \mathrm{s}$ The electric force is the only force acting on the electron. (a) Did the electron move to a higher potential or a lower potential? (b) Across what potential difference did the electron travel?
A hydrogen atom has a single proton at its center and a single electron at a distance of approximately \(0.0529 \mathrm{nm}\) from the proton. (a) What is the electric potential energy in joules? (b) What is the significance of the sign of the answer?
A defibrillator is used to restart a person's heart after it stops beating. Energy is delivered to the heart by discharging a capacitor through the body tissues near the heart. If the capacitance of the defibrillator is $9 \mu \mathrm{F}\( and the energy delivered is to be \)300 \mathrm{J},$ to what potential difference must the capacitor be charged?
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