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Chapter 20: Problem 47

A proton is at the origin and an ion is at \(x=5.0 \mathrm{nm}\). If the electric field is zero at \(x=-5\) nm, what's the ion's charge?

Short Answer

Expert verified
The ion's charge is \(0.4 \times 10^{-19} \, \text{C}\).

Step by step solution

01

Setup

Let's represent the charge of the ion as \(Q\). The distance from the ion to the point where the electric field is zero is \(5 + 5 = 10\) nm. The distance from the proton to the same point is 5 nm. Due to Coulomb's Law, the electric field at a point is the sum of the fields due to each charge, hence the electric fields due to the proton and the ion must cancel at -5 nm.
02

Expression for Electric Field

The electric field \(\text{E}\) due to a point charge \(q\) at a distance \(r\) from the charge is given by \[\text{E} = \frac{kq}{r^2}\] where \(k = 9.0 \times 10^9 \, \text{N m}^2/\text{C}^2\), the Coulomb's constant. Consequently, the field due to the proton would be \[\text{E}_p = \frac{k(1.6 \times 10^{-19}\, \text{C})}{(5 \, \text{nm})^2}\] and the field due to the ion would be \[\text{E}_i = \frac{kQ}{(10\, \text{nm})^2}\].
03

Equating Electric Fields

Since the electric field is zero at \(x = -5 \, \text{nm}\), we can set \(\text{E}_p = \text{E}_i\). Solving this equation gives us the charge of the ion \(Q\) as \[Q = 1.6 \times 10^{-19} \times \left(\frac{5}{10}\right)^2 = 0.4 \times 10^{-19} \, \text{C}\].

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Most popular questions from this chapter

You're \(1.5 \mathrm{m}\) from a charge distribution whose size is much less than \(1 \mathrm{m}\). You measure an electric field strength of \(282 \mathrm{N} / \mathrm{C}\) You move to a distance of \(2.0 \mathrm{m},\) and the field strength becomes 119 N/C. What's the net charge of the distribution? (Hint: Don't try to calculate the charge. Determine instead how the field decreases with distance, and from that infer the charge.)

Protons and neutrons are made from combinations of the two most common quarks, the \(u\) quark (charge \(+\frac{2}{3} e\) ) and the \(d\) quark (charge \(-\frac{1}{3} e\) ). How could three of these quarks combine to make (a) a proton and (b) a neutron?

Two identical small metal spheres initially carry charges \(q_{1}\) and \(q_{2} .\) When they're \(1.0 \mathrm{m}\) apart, they experience a \(2.5-\mathrm{N}\) attractive force. Then they're brought together so charge moves from one to the other until they have the same net charge. They're again placed \(1.0 \mathrm{m}\) apart, and now they repel with a \(2.5-\mathrm{N}\) force. What were the original charges \(q_{1}\) and \(q_{2} ?\)

You're taking physical chemistry, and your professor is discussing molecular dipole moments. Water, he says, has a dipole moment of "1.85 debyes," while carbon monoxide's dipole moment is only "0.12 debye." Your physics professor wants these moments expressed in SI. She tells you that the atomic separation in these two covalent compounds is about the same, and asks what that indicates about the way shared charge is distributed. What do you tell her?

The electron in a hydrogen atom is \(52.9 \mathrm{pm}\) from the proton. What's the proton's electric field strength at this distance?
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