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Chapter 22: Problem 38

What's the charge on an ion that gains \(1.6 \times 10^{-15} \mathrm{J}\) when it moves through a potential difference of \(2500 \mathrm{V} ?\)

Short Answer

Expert verified
The charge on an ion that gains \(1.6 \times 10^{-15} \mathrm{J}\) when it moves through a potential difference of \(2500 \mathrm{V}\) is \(6.4 \times 10^{-19}\) Coulombs.

Step by step solution

01

Identify Known Variables

From the exercise, we know that the energy or work done (W) is \(1.6 \times 10^{-15} \mathrm{J}\) and the potential difference (V) is \(2500 \mathrm{V}\).
02

Apply the Formula

The formula relating work done, charge and potential difference is \[W = QV\] where \(Q\) is the charge that needs to be found in this exercise. This formula allows us to find the charge \(Q\) by rearranging it as follows: \[Q = \frac{W}{V}\].
03

Substitute the Known Values

Substitute \(W\) with \(1.6 \times 10^{-15} \mathrm{J}\) and \(V\) with \(2500 \mathrm{V}\), which gives: \[Q = \frac{1.6 \times 10^{-15} \mathrm{J}}{2500 \mathrm{V}}\].
04

Solve

Solving the above expression would give us \(Q = 6.4 \times 10^{-19}\) Coulombs.

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Most popular questions from this chapter

An open-ended cylinder of radius \(a\) and length \(2 a\) carries charge \(q\) spread uniformly over its surface. Find the potential at the center of the cylinder. (Hint: Treat the cylinder as a stack of charged rings, and integrate.)

The potential difference between the surface of a 3.0 -cm-diameter power line and a point \(1.0 \mathrm{m}\) distant is \(3.9 \mathrm{kV}\). Find the line charge density on the power line.

A proton, an alpha particle (a bare helium nucleus), and a singly ionized helium atom are accelerated through a \(100-\mathrm{V}\) potential difference. How much energy does each gain?

A sphere of radius \(R\) carries negative charge of magnitude \(Q\), distributed in a spherically symmetric way. Find the escape speed for a proton at the sphere's surface that is, the speed that would enable the proton to escape to arbitrarily large distances starting at the sphere's surface.

A thin ring of radius \(R\) carries charge \(3 Q\) distributed uniformly over three-fourths of its circumference, and \(-Q\) over the rest. Find the potential at the ring's center.
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