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Chapter 24: Problem 30

Two concentric metal spheres are found to have a potential difference of \(900 . \mathrm{V}\) when a charge of \(6.726 \cdot 10^{-8} \mathrm{C}\) is applied to them. The radius of the outer sphere is \(0.210 \mathrm{~m}\). What is the radius of the inner sphere?

Short Answer

Expert verified
Answer: The radius of the inner sphere is approximately 0.141 m.

Step by step solution

01

Understand the given information and what we need to find

We are given the potential difference (\(V\)) between the two concentric metal spheres as \(900 \mathrm{V}\). We also know the charge (\(Q\)) applied to them is \(6.726 \cdot 10^{-8} \mathrm{C}\) and the radius of the outer sphere (\(R_2\)) is \(0.210 \mathrm{m}\). We need to find the radius of the inner sphere (\(R_1\)).
02

Use the potential difference formula

In this case, we will use the following formula to relate the potential difference, charge, and radii of the spheres: $$ V = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{\frac{1}{R_1} - \frac{1}{R_2}} $$ Where \(\epsilon_0\) is the vacuum permittivity constant and is equal to \(8.854 \times 10^{-12} \mathrm{F/m}\).
03

Rearrange the formula to solve for \(R_1\)

We want to find \(R_1\), so we will rearrange the given formula to solve for it: $$ \frac{1}{R_1} - \frac{1}{R_2} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{V} $$ $$ \frac{1}{R_1} = \frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{V} + \frac{1}{R_2} $$ $$ R_1 = \frac{1}{\frac{1}{4 \pi \epsilon_0} \cdot \frac{Q}{V} + \frac{1}{R_2}} $$
04

Calculate the radius of the inner sphere

Now we can plug in the values we are given to find \(R_1\): $$ R_1 = \frac{1}{\frac{1}{4 \pi (8.854 \times 10^{-12} \mathrm{F/m})} \cdot \frac{6.726 \times 10^{-8} \mathrm{C}}{900 \mathrm{V}} + \frac{1}{0.210 \mathrm{m}}} $$ Compute the value of \(R_1\): $$ R_1 \approx 0.141 \mathrm{m} $$ Therefore, the radius of the inner sphere is approximately \(0.141 \mathrm{m}\).

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Most popular questions from this chapter

How much energy can be stored in a capacitor with two parallel plates, each with an area of \(64.0 \mathrm{~cm}^{2}\) and separated by a gap of \(1.30 \mathrm{~mm}\), filled with porcelain whose dielectric constant is \(7.0,\) and holding equal and opposite charges of magnitude \(420 . \mu C ?\)

What is the radius of an isolated spherical conductor that has a capacitance of \(1.00 \mathrm{~F} ?\)

Consider a cylindrical capacitor, with outer radius \(R\) and cylinder separation \(d\). Determine what the capacitance approaches in the limit where \(d \ll R\). (Hint: Express the capacitance in terms of the ratio \(d / R\) and then examine what happens as the ratio \(d / R\) becomes very small compared to \(1 .)\) Explain why the limit on the capacitance makes sense.

A parallel plate capacitor of capacitance \(C\) is connected to a power supply that maintains a constant potential difference, \(V\). A close-fitting slab of dielectric, with dielectric constant \(\kappa\), is then inserted and fills the previously empty space between the plates. a) What was the energy stored on the capacitor before the insertion of the dielectric? b) What was the energy stored after the insertion of the dielectric? c) Was the dielectric pulled into the space between the plates, or did it have to be pushed in? Explain.

A typical AAA battery has stored energy of about 3400 J. (Battery capacity is typically listed as \(625 \mathrm{~mA} \mathrm{~h}\), meaning that much charge can be delivered at approximately \(1.5 \mathrm{~V}\).) Suppose you want to build a parallel plate capacitor to store this amount of energy, using a plate separation of \(1.0 \mathrm{~mm}\) and with air filling the space between the plates. a) Assuming that the potential difference across the capacitor is \(1.5 \mathrm{~V},\) what must the area of each plate be? b) Assuming that the potential difference across the capacitor is the maximum that can be applied without dielectric breakdown occurring, what must the area of each plate be? c) Is either capacitor a practical replacement for the AAA batterv?
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