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Chapter 17: Problem 69

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?

Short Answer

Expert verified
Answer: The distance between the spheres when the spark occurs is 0.3 mm or 3.0 x 10^-4 m.

Step by step solution

01

Find the electric field at the point where the spark occurs.

We can find the electric field at the point where the spark occurs by using the dielectric strength of dry air. The dielectric strength represents the maximum electric field that an insulating material, in this case, dry air, can withstand before breaking down. As a result, we can set the electric field at the point of the spark equal to the dielectric strength: \(E_{spark} = 3.0 \times 10^{6} \mathrm{\thinspace V/m}\)
02

Write down the formula for the electric field and potential difference.

The relationship between electric field and potential difference can be given as: \(V = E \times d\) Where: \(V\) is the potential difference between the two spheres, \(E\) is the electric field at a point in space, \(d\) is the distance between two points. We are given the potential difference between the spheres, \(V = 900 \mathrm{V}\), and we found the electric field at the point of the spark, \(E_{spark}\). Now we need to find the distance, \(d\), between the spheres at that point.
03

Solve for the distance between the spheres when the spark occurs.

Now let's substitute the values of the potential difference and the electric field into the formula and solve for the distance: \(900 \mathrm{V} = (3.0 \times 10^{6} \mathrm{V/m}) \times d\) To find the distance, we can divide both sides of the equation by \(3.0 \times 10^{6} \mathrm{V/m}\): \(d = \frac{900 \mathrm{V}}{3.0 \times 10^{6} \mathrm{V/m}}\) \(d = 3.0 \times 10^{-4} \mathrm{\thinspace m}\) So, the distance between the spheres when the spark occurs is \(3.0 \times 10^{-4} \mathrm{\thinspace m}\), or \(0.3 \mathrm{mm}\).

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Most popular questions from this chapter

(a) Calculate the capacitance per unit length of an axon of radius $5.0 \mu \mathrm{m} \text { (see Fig. } 17.14) .$ The membrane acts as an insulator between the conducting fluids inside and outside the neuron. The membrane is 6.0 nm thick and has a dielectric constant of \(7.0 .\) (Note: The membrane is thin compared with the radius of the axon, so the axon can be treated as a parallel plate capacitor.) (b) In its resting state (no signal being transmitted), the potential of the fluid inside is about \(85 \mathrm{mV}\) lower than the outside. Therefore, there must be small net charges \(\pm Q\) on either side of the membrane. Which side has positive charge? What is the magnitude of the charge density on the surfaces of the membrane?
An electron (charge \(-e\) ) is projected horizontally into the space between two oppositely charged parallel plates. The electric field between the plates is \(500.0 \mathrm{N} / \mathrm{C}\) upward. If the vertical deflection of the electron as it leaves the plates has magnitude \(3.0 \mathrm{mm},\) how much has its kinetic energy increased due to the electric field? [Hint: First find the potential difference through which the electron moves.]
The nucleus of a helium atom contains two protons that are approximately 1 fm apart. How much work must be done by an external agent to bring the two protons from an infinite separation to a separation of \(1.0 \mathrm{fm} ?\)
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