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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

A parallel plate capacitor is connected to a battery for charging. After some time, while the battery is still connected to the capacitor, the distance between the capacitor plates is doubled. Which of the following is (are) true? a) The electric field between the plates is halved. b) The potential difference of the battery is halved. c) The capacitance doubles. d) The potential difference across the plates does not change. e) The charge on the plates does not change.

An \(8.00-\mu F\) capacitor is fully charged by a \(240 .-V\) battery, which is then disconnected. Next, the capacitor is connected to an initially uncharged capacitor of capacitance \(C,\) and the potential difference across it is found to be \(80.0 \mathrm{~V}\) What is \(C ?\) How much energy ends up being stored in the second capacitor?

A \(4.00 \cdot 10^{3}-n F\) parallel plate capacitor is connected to a \(12.0-\mathrm{V}\) battery and charged. a) What is the charge \(Q\) on the positive plate of the capacitor? b) What is the electric potential energy stored in the capacitor? The \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor is then disconnected from the \(12.0-\mathrm{V}\) battery and used to charge three uncharged capacitors, a \(100 .-n F\) capacitor, a \(200 .-\mathrm{nF}\) capacitor, and a \(300 .-\mathrm{nF}\) capacitor, connected in series. c) After charging, what is the potential difference across each of the four capacitors? d) How much of the electrical energy stored in the \(4.00 \cdot 10^{3}-\mathrm{nF}\) capacitor was transferred to the other three capacitors?

Design a parallel plate capacitor with a capacitance of \(47.0 \mathrm{pF}\) and a capacity of \(7.50 \mathrm{nC}\). You have available conducting plates, which can be cut to any size, and Plexiglas sheets, which can be cut to any size and machined to any thickness. Plexiglas has a dielectric constant of 3.40 and a dielectric strength of \(4.00 \cdot 10^{7} \mathrm{~V} / \mathrm{m}\). You must make your capacitor as compact as possible. Specify all relevant dimensions. Ignore any fringe field at the edges of the capacitor plates.

A spherical capacitor is made from two thin concentric conducting shells. The inner shell has radius \(r_{1}\), and the outer shell has radius \(r_{2}\). What is the fractional difference in the capacitances of this spherical capacitor and a parallel plate capacitor made from plates that have the same area as the inner sphere and the same separation \(d=r_{2}-r_{1}\) between plates?
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