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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 the ion is \(6.4 \times 10^{-19} \mathrm{C}\).

Step by step solution

01

Understand the Given Information and the Required Variable

The energy \(U\) gained by the ion is \(1.6 \times 10^{-15} \mathrm{J}\) and the potential difference \(V\) it moved through is \(2500 \mathrm{V}\). We need to find the charge \(Q\) on the ion.
02

Use the Appropriate Formula

The formula relating potential difference, charge and potential energy is \(U = QV\). Rearrange this formula in order to solve for the unknown charge \(Q\), which gives \(Q = U/V\).
03

Substitute the Values into the Formula

Substitute the known values into the formula: \(Q = (1.6 \times 10^{-15} \mathrm{J}) / (2500 \mathrm{V})\).
04

Solve the Equation

Perform the division operation to find the value of the charge \(Q\).

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Most popular questions from this chapter

Find the magnitude of the potential difference between two points located \(1.4 \mathrm{m}\) apart in a uniform \(650-\mathrm{N} / \mathrm{C}\) electric field, if a line between the points is parallel to the field.

Show that \(1 \mathrm{V} / \mathrm{m}\) is the same as \(1 \mathrm{N} / \mathrm{C}\)

In considering the potential of an infinite flat sheet, why isn't it useful to take the zero of potential at infinity?

It takes \(45 \mathrm{J}\) to move a 15 -mC charge from point \(A\) to point \(B\) What's the potential difference \(\Delta V_{A B} ?\)

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