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What must be the distance between point charge \(q_{1}=\) \(26.0 \mu \mathrm{C}\) and point charge \(q_{2}=-47.0 \mu \mathrm{C}\) for the electrostatic force between them to have a magnitude of \(5.70 \mathrm{~N}\) ?

Short Answer

Expert verified
The required distance is approximately 0.0438 meters.

Step by step solution

01

Understand the Problem

We have two point charges: \(q_1 = 26.0 \mu C\) and \(q_2 = -47.0 \mu C\). We need to find the distance \(r\) such that the electrostatic force \(F\) between them equals \(5.70 \mathrm{~N}\).
02

Use Coulomb’s Law

The electrostatic force between two point charges is given by Coulomb's Law: \[F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\] where \(k\) is Coulomb's constant, \(8.99 \times 10^9 \mathrm{~Nm}^2/\mathrm{C}^2\), \(|q_1|\) and \(|q_2|\) are the magnitudes of the charges, and \(r\) is the distance between them.
03

Substitute Known Values into Coulomb’s Law

Substitute the known values into the formula: \[5.70 = \frac{8.99 \times 10^9 \times 26.0 \times 10^{-6} \times 47.0 \times 10^{-6}}{r^2}\].
04

Solve for Distance \(r\)

Rearrange the equation to solve for \(r^2\): \[r^2 = \frac{8.99 \times 10^9 \times 26.0 \times 10^{-6} \times 47.0 \times 10^{-6}}{5.70}\].Calculate the right side and then take the square root to find \(r\): \[r = \sqrt{\frac{8.99 \times 10^9 \times 26.0 \times 10^{-6} \times 47.0 \times 10^{-6}}{5.70}}\].
05

Compute the Final Result

Calculate the numerical values: \[r^2 \approx \frac{10.955562 \times 10^9 \times 10^{-12}}{5.70} \approx \frac{10.955562 \times 10^{-3}}{5.70} \approx \frac{0.010955562}{5.70}\],\[r^2 \approx 0.0019220\].Taking the square root, \(r \approx \sqrt{0.0019220} \approx 0.0438\) meters.

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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 the force of attraction or repulsion between two charged objects. This force plays a fundamental role in the world of physics and is vital to understanding interactions at atomic and molecular levels.
The magnitude of electrostatic force is determined using Coulomb's Law, which provides a way to calculate how strong the force is between two charges. The formula is given by: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} \]where:
  • \(F\) is the force between the charges.
  • \(k\) is Coulomb's constant, approximately \(8.99 \times 10^9 \, \mathrm{Nm}^2/\mathrm{C}^2\).
  • \(|q_1|\) and \(|q_2|\) are the magnitudes of the point charges.
  • \(r\) is the distance between the charges.
Through this equation, we can understand that:
  • If the charges are of the same sign, the force will be repulsive.
  • If the charges are of opposite signs, the force will be attractive.
The electrostatic force decreases with the square of the distance, which means as the distance between charges increases, the force decreases rapidly. This concept is crucial for understanding how charged objects interact both in close proximity and from afar.
Point Charges
Point charges are idealized models where the size of a charged object is assumed to be infinitesimally small. This simplification helps in focusing on the charge's effects without considering the object's shape or size.
In practical terms, a point charge can represent:
  • Tiny charged particles like electrons or protons.
  • Larger objects where the charge is evenly distributed over a surface or volume, which can be treated as a point for simplicity if you're calculating effects at a significant distance away.
The usefulness of the point charge model lies in its ability to simplify complex calculations, making it much easier to analyze electric fields and forces. For instance, when using Coulomb's Law to determine the force between charges, treating them as point charges allows for direct application of the formula, without needing to consider intricacies of the object's size or shape.
Distance Calculation
Calculating the distance between two charges is often necessary to determine the magnitude of electrostatic force using Coulomb's Law. The distance is a critical factor because it affects the strength of the force exponentially.
Let's see why distance matters:
  • The formula \(F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\) shows that force is inversely proportional to the square of the distance \(r\). This inverse square law means that as the distance doubles, the force reduces to a fourth of its value.
  • To solve for the distance, rearrange the formula to get \(r\): \[ r = \sqrt{\frac{k \cdot |q_1 \cdot q_2|}{F}} \]
When calculating this for our exercise's point charges, substitute known values for the charges and the desired force to directly solve for \(r\). In practice, this involves evaluating the expression, performing arithmetic calculations, and ultimately solving for \(r\) by taking the square root.Distance is measurable and can be evaluated through calculations or experimentally, providing both theoretical and practical understanding of electrostatic interactions.

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Most popular questions from this chapter

Shows four identical conducting spheres that are actually well separated from one another. Sphere \(W\) (with an initial charge of zero) is touched to sphere \(A\) and then they are separated. Next, sphere \(W\) is touched to sphere \(B\) (with an initial charge of \(-32 e\) ) and then they are separated. Finally, sphere \(W\) is touched to sphere \(C\) (with an initial charge of \(+48 e\) ), and then they are separated. The final charge on sphere \(W\) is \(+18 e\). What was the initial charge on sphere \(A\) ?

Point charges of \(+6.0 \mu \mathrm{C}\) and \(-4.0 \mu \mathrm{C}\) are placed on an \(x\) axis, at \(x=8.0 \mathrm{~m}\) and \(x=16 \mathrm{~m}\), respectively. What charge must be placed at \(x=24 \mathrm{~m}\) so that any charge placed at the origin would experience no electrostatic force?

A current of \(0.300\) A through your chest can send your heart into fibrillation, ruining the normal rhythm of heartbeat and disrupting the flow of blood (and thus oxygen) to your brain. If that current persists for \(2.00 \mathrm{~min}\), how many conduction electrons pass through your chest?

A neutron consists of one "up" quark of charge \(+2 e / 3\) and two "down" quarks each having charge \(-e / 3 .\) If we assume that the down quarks are \(2.6 \times 10^{-15} \mathrm{~m}\) apart inside the neutron, what is the magnitude of the electrostatic force between them?

The magnitude of the electrostatic force between two identical ions that are separated by a distance of \(5.0 \times 10^{-10} \mathrm{~m}\) is \(3.7 \times 10^{-9} \mathrm{~N}\). (a) What is the charge of each ion? (b) How many electrons are "missing" from each ion (thus giving the ion its charge imbalance)?

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