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The force between a charge of \(25 \mu C\) and a charge of \(-10 \mu C\) is \(8.0 \mathrm{~N}\). What is the separation between the two charges? a) \(0.28 \mathrm{~m}\) c) \(0.45 \mathrm{~m}\) b) \(0.53 \mathrm{~m}\) d) \(0.15 \mathrm{~m}\)

Short Answer

Expert verified
(a) 0.5 m (b) 0.53 m (c) 0.56 m (d) 0.6 m Answer: (b) 0.53 m

Step by step solution

01

Write down the given information

We are given the force between the charges \((F = 8.0 \mathrm{~N})\), the charges \((q_1 = 25 \times 10^{-6} C \textrm{ and } q_2 = -10 \times 10^{-6} C)\), and the distances \((a, b, c, d)\). We can now use the information to solve the problem using Coulomb's law.
02

Use Coulomb's law to find the distance between the charges

We can rearrange the formula for Coulomb's law to solve for the distance between the charges: \(r = \sqrt{\frac{k \cdot q_1 \cdot q_2}{F}}\). Now, we can plug in the given values into the formula: \(r = \sqrt{\frac{8.9875 \times 10^{9} \cdot 25 \times 10^{-6} \cdot (-10 \times 10^{-6})}{8.0}}\)
03

Calculate the distance

By performing the calculations, we get: \(r = \sqrt{\frac{-2.246875 \times 10^{4}}{8}}\) \(r = \sqrt{-2810.859375}\) Since force F is in N, which is a positive value and charge q_1*q_2 is negative the equation is not correct in terms of units. The correct formula should be: \(r = \sqrt{\frac{k \cdot |q_1 \cdot q_2|}{F}}\) \(r = \sqrt{\frac{8.9875 \times 10^{9} \cdot |25 \times 10^{-6} \cdot (-10 \times 10^{-6})|}{8.0}}\) \(r = \sqrt{0.28125}\) \(r ≈ 0.53\mathrm{~m}\) Thus, the separation between the two charges is approximately \(0.53\mathrm{~m}\), which corresponds to option (b).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electric Charges
Electric charges are the foundation of electrostatics. They come in two types: positive and negative. Like charges repel each other while opposite charges attract. This basic property of attraction and repulsion between charges explains a multitude of phenomena, from the behavior of atoms to electrical technology.

Objects can be charged in several ways, including by friction, which transfers charge from one object to another, or by induction, where a charged object brought near a neutral one causes a separation of charges within the neutral object. A key characteristic of charge, denoted by the symbol 'q', is that it is conserved. In any system, the total charge remains constant, and this principle applies to the interactions between charges as well.
Electrostatic Force
Electrostatic force is the force that electric charges exert on each other. This force can act over a distance, allowing charged objects to interact without directly touching. It's one of the fundamental forces in physics and is governed by Coulomb's Law.

The electrostatic force is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between their centers. It is immensely powerful on a small scale and is responsible for holding atoms and molecules together. However, as the distance between charges increases, the force decreases sharply, which is why we don’t feel these forces in our everyday life.
Coulomb's Law Formula
Coulomb's Law formula quantitatively describes the electrostatic force between two charges. The law was named after Charles-Augustin de Coulomb, who introduced the formula in the 18th century. The formula is as follows:
\[F = k\frac{|q_1 \times q_2|}{r^2}\]
Here, \(F\) represents the force between charges, \(k\) is the Coulomb's constant (\(8.99 \times 10^9 \mathrm{N m^2/C^2}\)), \(q_1\) and \(q_2\) are the magnitudes of the charges, and \(r\) is the distance between the centers of the two charges. The absolute value is used because the force is always positive, indicating it has a magnitude regardless of the charges being like or opposite.

It's crucial to use the distance \(r\) accurately as it greatly affects the calculated force—doubling the distance decreases the force by a factor of four, illustrating the inverse square nature of the relationship.

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Most popular questions from this chapter

Two positive charges, each equal to \(Q,\) are placed a distance \(2 d\) apart. A third charge, \(-0.2 Q\), is placed exactly halfway between the two positive charges and is displaced a distance \(x \ll d\) (that is, \(x\) is much smaller than \(d\) ) perpendicular to the line connecting the positive charges. What is the force on this charge? For \(x \ll d\), how can you approximate the motion of the negative charge?

Your sister wants to participate in the yearly science fair at her high school and asks you to suggest some exciting project. You suggest that she experiment with your recently created electron extractor to suspend her cat in the air. You tell her to buy a copper plate and bolt it to the ceiling in her room and then use your electron extractor to transfer electrons from the plate to the cat. If the cat weighs \(7.00 \mathrm{~kg}\) and is suspended \(2.00 \mathrm{~m}\) below the ceiling, how many electrons have to be extracted from the cat? Assume that the cat and the metal plate are point charges.

In general, astronomical objects are not exactly electrically neutral. Suppose the Earth and the Moon each carry a charge of \(-1.00 \cdot 10^{6} \mathrm{C}\) (this is approximately correct; a more precise value is identified in Chapter 22 ). a) Compare the resulting electrostatic repulsion with the gravitational attraction between the Moon and the Earth. Look up any necessary data. b) What effects does this electrostatic force have on the size, shape, and stability of the Moon's orbit around the Earth?

How is it possible for one electrically neutral atom to exert an electrostatic force on another electrically neutral atom?

A silicon sample is doped with phosphorus at 1 part per \(1.00 \cdot 10^{6}\) Phosphorus acts as an electron donor, providing one free electron per atom. The density of silicon is \(2.33 \mathrm{~g} / \mathrm{cm}^{3},\) and its atomic mass is \(28.09 \mathrm{~g} / \mathrm{mol}\). a) Calculate the number of free (conduction) electrons per unit volume of the doped silicon. b) Compare the result from part (a) with the number of conduction electrons per unit volume of copper wire, assuming that each copper atom produces one free (conduction) electron. The density of copper is \(8.96 \mathrm{~g} / \mathrm{cm}^{3}\), and its atomic mass is \(63.54 \mathrm{~g} / \mathrm{mol}\).

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