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When point chargcs \(q_{1}=+8.4 \mu \mathrm{C}\) and \(q_{2}=+5.6 \mu \mathrm{C}\) are brought near each other, each experiences a repulsive force of magnitude 0.66 \(\mathrm{N}\) . Determine the distance between the charges.

Short Answer

Expert verified
The distance between the charges is approximately 0.800 meters.

Step by step solution

01

Understand Given Variables and Coulomb's Law

First, identify the values given in the problem: \( q_1 = +8.4 \mu C = 8.4 \times 10^{-6} C \), \( q_2 = +5.6 \mu C = 5.6 \times 10^{-6} C \), and the force \( F = 0.66 \text{ N} \). We will use Coulomb's Law to find the distance. Coulomb’s law: \( F = k \frac{|q_1 q_2|}{r^2} \), where \( k = 8.99 \times 10^9 \text{Nm}^2/\text{C}^2 \) is Coulomb's constant.
02

Rearrange Coulomb's Law

Our goal is to solve for the distance \( r \) between the two charges. Rearrange the Coulomb's law formula to isolate \( r \): \[ r^2 = k \frac{|q_1 q_2|}{F} \]\[ r = \sqrt{k \frac{|q_1 q_2|}{F}} \]
03

Substitute Known Values into the Equation

Substitute the known values into the formula:\[ r = \sqrt{(8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2) \left(\frac{(8.4 \times 10^{-6} \text{ C}) \cdot (5.6 \times 10^{-6} \text{ C})}{0.66 \text{ N}}\right)} \]
04

Calculate to Find the Distance

Perform the calculations:\[ r = \sqrt{(8.99 \times 10^9) \times \left(\frac{47.04 \times 10^{-12}}{0.66}\right)} \]\[ r = \sqrt{(8.99 \times 10^9) \times (71.273 \times 10^{-12})} \]\[ r = \sqrt{640.283 \times 10^{-3}} \]Finally, calculate:\[ r \approx 0.800 \text{ m} \]
05

Conclusion: Distance Between the Charges

The distance between the two point charges is approximately 0.800 meters.

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Key Concepts

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

Electric Force
Electric force is the push or pull that occurs between charged objects. It is an essential concept in physics that describes how objects with electric charges interact with each other.
The electric force can be either attractive or repulsive. When both objects have charges of the same type, like both positive or both negative, they repel each other. Conversely, if they have opposite charges, such as one positive and one negative, they attract each other. This behavior is similar to magnets, where like poles repel and opposite poles attract.
  • Electric force is a vector quantity, meaning it has both magnitude and direction.
  • It is expressed in Newtons (N), just like any other force.
  • This force becomes weaker as the distance between the charges increases.
To quantify this force, we use the principle known as Coulomb's Law, which allows us to calculate the magnitude of the force based on the size of the charges and the distance between them. Understanding this concept is crucial for solving problems involving electric interactions.
Point Charges
Point charges are idealized charges that are treated as if they occupy a single point in space. These are theoretical constructs used to simplify complex calculations and provide a clearer picture of electric interactions. Because real charges are often distributed over space, thinking of them as point charges can make analysis more straightforward.
  • Point charges help to simplify electrostatic scenarios by focusing solely on the interaction between charges.
  • They allow the application of Coulomb’s Law straightforwardly, considering only the magnitudes of the charges and the distance between them.
  • These are usually small or compact charges where size is negligible.
For example, when calculating the electric force between two charged particles in an exercise, you often treat these particles as point charges. This means you don’t factor in their dimension but rather only their charge and distance. This simplification is very useful in understanding basic physics problems regarding electrostatics, like the one solved in the original exercise.
Calculating Distance Between Charges
To calculate the distance between two point charges, we often resort to Coulomb’s Law, which is a key formula in electrostatics. Coulomb's Law provides a mathematical framework that defines how the electric force relates to the distance between two charged entities:
\[ F = k \frac{|q_1 q_2|}{r^2} \]\
Where:
  • \( F \) is the magnitude of the electric force.
  • \( k \) is Coulomb's constant, approximately equal to 8.99 × 109 ±·³¾Â²/°ä².
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  • \( r \) is the distance between the centers of the two charges.
By rearranging this formula, we can solve for the distance \( r \) when the force and both charges are known. It involves isolating \( r \) on one side of the equation:\[ r = \sqrt{k \frac{|q_1 q_2|}{F}} \]
This rearrangement lets you plug in known values for \( q_1 \), \( q_2 \), and \( F \) to compute \( r \), helping determine how far apart the charges are in physical space.

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Most popular questions from this chapter

Two point chargcs are located along the \(x\) axis: \(q_{1}=+6.0 \mu \mathrm{C}\) at \(x_{1}=+4.0 \mathrm{cm},\) and \(q_{2}=+6.0 \mu \mathrm{C}\) at \(x_{2}=-4.0 \mathrm{cm} .\) Two other charges are located on the y axis: \(q_{3}=+3.0 \mu \mathrm{C}\) at \(y_{1}=+5.0 \mathrm{cm},\) and \(q_{4}=-8.0 \mu \mathrm{C}\) at \(y_{4}=+7.0 \mathrm{cm} .\) Find the net clectric ficld (magnitude and direction) at the origin.

Four point charges have equal magnitudes. Three are positive, and one is negative, as the drawing shows. They are fixed in place on the same straight line, and adjacent charges are equally separated by a distance d. Consider the net electrostatic force acting on each charge. Calculate the ratio of the largest to the smallest net force.

A cube is located with one corner situated at the origin of an x, y, z coordinate system. One of the cube’s faces lies in the x, y plane, another in the y, z plane, and another in the x, z plane. In other words, the cube is in the first octant of the coordinate system. The edges of the cube are 0.20 m long. A uniform electric field is parallel to the x, y plane and points in the direction of the y axis. The magnitude of the field is 1500 N/C. (a) Using the outward normal for each face of the cube, find the electric flux through each of the six faces. (b) Add the six values obtained in part (a) to show that the electric flux through the cubical surface is zero, as Gauss’ law predicts, since there is no net charge within the cube.

A charge \(Q\) is located inside a rectangular box. The electric flux through each of the six surfaces of the box is: \(\Phi_{1}=+1500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\) \(\Phi_{2}=+2200 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{3}=+4600 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}, \Phi_{4}=-1800 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}\) \(\Phi_{5}=-3500 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C},\) and \(\Phi_{6}=-5400 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} .\) What is \(Q ?\)

In a vacuum, two particles have charges of \(q_{1}\) and \(q_{2},\) where \(q_{1}=+3.5 \mu \mathrm{C}\) They are separated by a distance of \(0.26 \mathrm{m},\) and particle l experiences an allractive force of 3.4 \(\mathrm{N}\) . What is \(q_{2}\) (magnitude and sign)?

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