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A \(7.5-n C\) charge is located \(1.8 \mathrm{~m}\) from a \(4.2-n C\) charge. Find the magnitude of the electrostatic force that one charge exerts on the other. Is the force attractive or repulsive?

Short Answer

Expert verified
The force that one charge exerts on the other is approximately \(0.0875 N\), and it is repulsive.

Step by step solution

01

Identify Known Variables

The known variables are the magnitudes of the charges, \(q_1 = 7.5 nC\) and \(q_2 = 4.2 nC\), and the distance between the charges, \(r = 1.8 m\). The values must be used in the SI unit of charge, Coulomb. Therefore, \(q_1 = 7.5 × 10^{-9} C\) and \(q_2 = 4.2 × 10^{-9} C\).
02

Apply Coulomb's Law

Coulomb's law provides the magnitude of the force (\(F\)) between two charges: \( F = k * \frac{{|q1 * q2|}}{{r^2}} \). Here, \(k\) is Coulomb's constant (\(8.99 × 10^9 Nm^2/C^2\)). Substituting the known values, \( F = 8.99 × 10^9 * \frac{{(7.5 × 10^{-9})(4.2 × 10^{-9})}}{{(1.8)^2}} \).
03

Calculate the Force

Calculate the force that one charge exerts on the other. After the necessary calculations, F ≈ 0.0875 N.
04

Determine the Nature of the Force

Since both charges are positive, they will repel each other. Therefore, the force is repulsive.

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Key Concepts

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

Electrostatic Force
Electrostatic force is a fundamental concept in physics, especially important in understanding how charged particles interact. It is the force that acts between charged objects. When charges are stationary, the force governing their interactions is known as electrostatic force. Coulomb's Law is used to calculate the magnitude of this force. It states that the force between two point charges is directly proportional to the product of the absolute values of the charges and inversely proportional to the square of the distance between them.

In formula terms, Coulomb's Law is expressed as:
  • \( F = k * \frac{{|q_1 * q_2|}}{{r^2}} \)
where:
  • \( F \) is the electrostatic force.
  • \( k \) is Coulomb's constant, approximately \(8.99 \times 10^9 \text{Nm}^2/ ext{C}^2\).
  • \( q_1 \) and \( q_2 \) are the point charges.
  • \( r \) is the distance between the charges.
Understanding this principle helps explain a wide range of phenomena, from the behavior of atoms and molecules to the functioning of larger systems like capacitors in electronic circuits.
Charge Interactions
The way charges interact is foundational to many branches of physics and everyday phenomena. Charges come in two types: positive and negative. The interactions between charges are determined by their signs:
  • Like charges (both positive or both negative) repel each other. The force pushes the charges apart.
  • Opposite charges (one positive, one negative) attract each other. Here, the force pulls the charges together.
In our exercise, the force is repulsive because both given charges are positive. This means that each charge experiences a force pushing it away from the other. This interaction helps us understand why like charges don't accumulate easily in one place and how they seek to minimize energy states by distancing themselves.
SI Units
The International System of Units (SI Units) is crucial for consistent scientific communication. In the context of electrostatics, the charge is measured in Coulombs, which is the standard SI unit for electric charge. However, in many practical contexts, charges are often given in smaller units such as nanocoulombs (nC) because typical everyday charges are much smaller than one Coulomb.
  • 1 Coulomb (C) is equivalent to \(10^9\) nanocoulombs (nC).
In the original exercise, charges were provided in nanocoulombs. Converting them to Coulombs is essential when performing calculations to ensure accuracy and standardization. A uniform unit system, like the SI system, allows scientists to ensure measurements and calculations are interpretable and comparable across different studies and fields.

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Most popular questions from this chapter

A solid conducting sphere of radius \(2.00 \mathrm{~cm}\) has a charge of \(8.00 \mu \mathrm{C}\). A conducting spherical shell of inner radius \(4.00 \mathrm{~cm}\) and outer radius \(5.00 \mathrm{~cm}\) is concentric with the solid sphere and has a charge of \(-4.00 \mu \mathrm{C}\). Find the electric field at (a) \(r=1.00 \mathrm{~cm}\), (b) \(r=3.00 \mathrm{~cm}\), (c) \(r=4.50 \mathrm{~cm}\), and (d) \(r=7.00 \mathrm{~cm}\) from the center of this charge configuration.

Protons are projected with an initial speed \(v_{0}=9550 \mathrm{~m} / \mathrm{s}\) into a region where a uniform electric field \(E=720 \mathrm{~N} / \mathrm{C}\) is present (Fig. P15.64). The protons are to hit a target that lies a horizontal distance of \(1.27 \mathrm{~mm}\) from the point where the protons are launched. Find (a) the two projection angles \(\theta\) that will result in a hit and (b) the total duration of flight for each of the two trajectories.

Each of the electrons in a particle beam has a kinetic energy of \(1.60 \times 10^{-17} \mathrm{~J}\). (a) What is the magnitude of the uniform electric field (pointing in the direction of the electrons' movement) that will stop these electrons in a distance of \(10.0 \mathrm{~cm}\) ? (b) How long will it take to stop the electrons? (c) After the electrons stop, what will they do? Explain.

A charge of \(-5.0 \mathrm{nC}\) is at the origin and a second charge of \(7.0 \mathrm{nC}\) is at \(x=4.00 \mathrm{~m}\). Find the magnitude and direction of the clectric ficld halfway in between the two charges.

Each of the protons in a particle beam has a kinetic energy of \(3.25 \times 10^{-15} \mathrm{~J}\). What are the magnitude and direction of the electric field that will stop these protons in a distance of \(1.25 \mathrm{~m}\) ?

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