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Two small, positively charged spheres have a combined charge of \(5.0 \times 10^{-5} \mathrm{C}\). If each sphere is repelled from the other by an electrostatic force of \(1.0 \mathrm{~N}\) when the spheres are \(2.0 \mathrm{~m}\) apart, what is the charge on the sphere with the smaller charge?

Short Answer

Expert verified
The charge on the sphere with the smaller charge is approximately \(2.43 \times 10^{-5} \mathrm{~C}\).

Step by step solution

01

Understand Coulomb's Law

Coulomb's Law describes the electrostatic force between two charges. It is given by the formula: \[ F = k \frac{q_1 q_2}{r^2} \]where \( F \) is the force, \( k = 8.99 \times 10^9 \mathrm{~Nm^2/C^2} \) is Coulomb's constant, \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between the charges.
02

Set up the Equation Using Given Values

The force \( F = 1.0 \mathrm{~N} \), distance \( r = 2.0 \mathrm{~m} \), and the total charge \( q_1 + q_2 = 5.0 \times 10^{-5} \mathrm{~C} \). Substitute these into Coulomb's formula: \[ 1.0 = 8.99 \times 10^9 \frac{q_1 q_2}{(2.0)^2} \]Simplify it to: \[ q_1 q_2 = \frac{1.0 \times 4}{8.99 \times 10^9} \]
03

Simplify and Solve for Product of Charges

Calculate the product \( q_1 q_2 \):\[ q_1 q_2 = \frac{4}{8.99 \times 10^9} \approx 4.45 \times 10^{-10} \mathrm{~C^2} \]
04

Solve the System of Equations

Now, solve the system of equations: 1. \( q_1 + q_2 = 5.0 \times 10^{-5} \)2. \( q_1 q_2 = 4.45 \times 10^{-10} \).Use the quadratic formula where \( q_1 \) and \( q_2 \) are the roots: \[ x^2 - (5.0 \times 10^{-5})x + 4.45 \times 10^{-10} = 0 \]
05

Apply the Quadratic Formula

Solve the quadratic equation using the quadratic formula: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]where \( a = 1 \), \( b = -5.0 \times 10^{-5} \), and \( c = 4.45 \times 10^{-10} \).
06

Calculate the Charges

Using the quadratic formula:\[ x = \frac{5.0 \times 10^{-5} \pm \sqrt{(5.0 \times 10^{-5})^2 - 4 \times 4.45 \times 10^{-10}}}{2} \]Calculate: \[ x \approx \frac{5.0 \times 10^{-5} \pm 1.15 \times 10^{-6}}{2} \]Thus, the charges are approximately:\[ q_1 \approx 2.43 \times 10^{-5} \mathrm{~C} \] and \[ q_2 \approx 2.57 \times 10^{-5} \mathrm{~C} \].
07

Identify the Smaller Charge

The charges calculated are \( q_1 = 2.43 \times 10^{-5} \mathrm{~C} \) and \( q_2 = 2.57 \times 10^{-5} \mathrm{~C} \). The smaller charge is \( 2.43 \times 10^{-5} \mathrm{~C} \).

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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, particularly when dealing with charged particles or objects. It refers to the force of attraction or repulsion between two charged bodies. According to Coulomb's Law, this force, denoted by \( F \), can be calculated using the equation: \[ F = k \frac{q_1 q_2}{r^2} \] Where:
  • \( F \) is the electrostatic force.
  • \( k \) is Coulomb's constant (\(8.99 \times 10^9 \, \mathrm{Nm^2/C^2} \)).
  • \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
  • \( r \) is the distance between the centers of the two charges.
Coulomb's Law is essential for predicting the behavior of charged particles. An important thing to note is that the force is inversely proportional to the square of the distance between charges. This means if the distance doubles, the force becomes one-quarter as strong, and if it halves, the force increases fourfold. Understanding this relationship is crucial for problems involving electrostatic force calculations.
Quadratic Formula
In problems involving charges, often the quadratic formula is used to find unknown values related to the charges, especially when the problem involves the product of two charges and their sum. The quadratic equation is given in the form: \[ ax^2 + bx + c = 0 \] To solve for \( x \), the quadratic formula is: \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] This formula requires finding the values of \( a \), \( b \), and \( c \) from the quadratic equation. The plus-minus symbol (\( \pm \)) suggests there are usually two possible solutions for \( x \). In the context of charge problems, it helps find the two possible values of charges, \( q_1 \) and \( q_2 \). Ensuring the correct setup of the quadratic equation is vital. This accuracy affects whether the roots obtained are meaningful in the real-world context of charges. The discriminant \( (b^2 - 4ac) \) helps determine the number and type of solutions: if positive, two real roots; if zero, one real root; and if negative, no real roots within the realm of real numbers.
Charge Calculation
Charge calculation in physics often involves utilizing relationships between the charges, such as their sum and product. Consider a scenario where the sum of two charges is given, and their interaction force is known at a specific distance. By using Coulomb's Law and considering the given total charge, we have:
  • Total charge: \( q_1 + q_2 = C_t \)
  • Product of charges from Coulomb's Law: \( q_1 q_2 = F \times \frac{r^2}{k} \)
These two expressions form a system of equations solvable using the quadratic formula. The values plugged into the quadratic formula reveal the individual charges. Engaging with such calculations requires understanding and accurately implementing algebraic manipulation and quadratic solving techniques. Take care when simplifying or substituting numerical values to avoid computational errors. Particularly in problems involving small or large numbers, which is common in electrostatics, where precision in handling powers of ten is crucial to accurately deducing the correct charges.

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Most popular questions from this chapter

A charged nonconducting rod, with a length of \(2.00 \mathrm{~m}\) and a cross- sectional area of \(4.00 \mathrm{~cm}^{2}\), lies along the positive side of an \(x\) axis with one end at the origin. The volume charge density \(\rho\) is charge per unit volume in coulombs per cubic meter. How many excess electrons are on the rod if \(\rho\) is (a) uniform, with a value of \(-4.00 \mu \mathrm{C} / \mathrm{m}^{3},\) and \((\mathrm{b})\) nonuniform, with a value given by \(\rho=b x^{2}\) where \(b=-2.00 \mu \mathrm{C} / \mathrm{m}^{5} ?\)

Earth’s atmosphere is constantly bombarded by cosmic ray protons that originate somewhere in space. If the protons all passed through the atmosphere, each square meter of Earth's surface would intercept protons at the average rate of 1500 protons per second. What would be the electric current intercepted by the total surface area of the planet?

An electron is projected with an initial speed \(v_{i}=3.2 \times 10^{5} \mathrm{~m} / \mathrm{s}\) directly toward a very distant proton that is at rest. Because the proton mass is large relative to the electron mass, assume that the proton remains at rest. By calculating the work done on the electron by the electrostatic force, determine the distance between the two particles when the electron instantaneously has speed \(2 v_{i}\).

A particle of charge \(+3.00 \times 10^{-6} \mathrm{C}\) is \(12.0 \mathrm{~cm}\) distant from a second particle of charge \(-1.50 \times 10^{-6} \mathrm{C}\). Calculate the magnitude of the electrostatic force between the particles.

Two tiny, spherical water drops, with identical charges of \(-1.00 \times 10^{-16} \mathrm{C},\) have a center-to-center separation of \(1.00 \mathrm{~cm} .\) (a) What is the magnitude of the electrostatic force acting between them? (b) How many excess electrons are on each drop, giving it its charge imbalance?

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