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Chapter 20: Problem 41

You have two charges \(+4 q\) and one charge \(-q .\) How would you place them along a line so there's no net force on any of the three?

Short Answer

Expert verified
The charges should be arranged in line as follows: +4q, -q, +4q to ensure that there will be no net force acting on any of the three charges.

Step by step solution

01

Analyze the charges

There are two charges of +4q and one charge of -q. As such, the charge -q will be attracted to the two +4q charges.
02

Arrangement of the charges

The negative charge should be positioned in between the two positive charges. This is because it is attracted to the positive charge and repelled by other negative charges. Therefore, the suitable arrangement starting from the left side would be +4q, -q, +4q.
03

Understanding the no net force condition

The no net force condition means that the sum of the forces acting on the individual charges must be zero. Here, for the two +4q charges, the net force will obviously be zero, as they both will feel the same amount of force due to the -q charge placed in between them. However, we need to be careful while considering the -q charge. The -q charge will have an equivalent amount of attractive force towards both the +4q charges.

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Most popular questions from this chapter

The ring in Example 20.6 carries total charge \(Q,\) and the point \(P\) is the same distance \(r=\sqrt{x^{2}+a^{2}}\) from all parts of the ring. So why isn't the electric field of the ring just \(k Q / r^{2} ?\)

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.)

Conceptual Example 20.1 shows that the gravitational force between an electron and a proton is about \(10^{-40}\) times weaker than the electric force between them. since matter consists largely of electrons and protons, why is the gravitational force important?

Show that the electric field at a point \(45^{\circ}\) from a dipole's axis is \(1 / \sqrt{5 / 8}\) times the field on the axis, assuming both points are the same distance from the dipole and that distance is large compared with the dipole spacing.

The electron and proton in a hydrogen atom are \(52.9 \mathrm{pm}\) apart. Find the magnitude of the electric force between them.
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