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Two small plastic spheres are given positive electric charges. When they are 15.0 cm apart, the repulsive force between them has magnitude 0.220 N. What is the charge on each sphere (a) if the two charges are equal and (b) if one sphere has four times the charge of the other?

Short Answer

Expert verified
(a) Each sphere has a charge of approximately \( 7.42 \times 10^{-6} \text{ C} \). (b) The smaller charge is \( 3.71 \times 10^{-6} \text{ C} \), and the larger charge is \( 1.48 \times 10^{-5} \text{ C} \).

Step by step solution

01

Understanding Coulomb's Law

The force exerted between two point charges is defined by Coulomb's law, given by the formula: \[ F = k \frac{|q_1 q_2|}{r^2} \] where \( F \) is the force between the charges, \( k \) is Coulomb's constant \((8.988 \times 10^9 \text{ Nm}^2/\text{C}^2)\), \( q_1 \) and \( q_2 \) are the magnitudes of the charges, and \( r \) is the distance between the charges.
02

Calculation for Equal Charges \( (q_1 = q_2 = q) \)

If the charges are equal, \( q_1 = q_2 = q \). Plug \( q_1 = q_2 = q \) and \( r = 0.15 \text{ m} \) into the formula: \[ 0.220 = (8.988 \times 10^9) \frac{q^2}{(0.15)^2} \]. Solving for \( q \), we find \[ q^2 = \frac{0.220 \times (0.15)^2}{8.988 \times 10^9} \]. Compute \( q \) by taking the square root of the right side of the equation.
03

Solving for \( q \): Equal Charges

Calculate \( q^2 \) which equals approximately \( 5.50 \times 10^{-12} \). Taking the square root, we find \( q \approx 7.42 \times 10^{-6} \text{ C} \). So each sphere has a charge of approximately \( 7.42 \times 10^{-6} \text{ C} \).
04

Understanding Unequal Charge Distribution

For case (b), where one charge is four times the other, denote \( q_1 = q \) and \( q_2 = 4q \). Substitute into the formula: \[ 0.220 = (8.988 \times 10^9) \frac{q \cdot 4q}{(0.15)^2} \].
05

Solving for \( q \): One Charge is Four Times the Other

The expression becomes \[ 0.220 = (8.988 \times 10^9) \frac{4q^2}{(0.15)^2} \]. Solving for \( q^2 \), we get \[ q^2 = \frac{0.220 \times (0.15)^2}{4 \times 8.988 \times 10^9} \]. Calculate \( q^2 \) and find \( q \approx 3.71 \times 10^{-6} \text{ C} \). Thus, \( q_1 = 3.71 \times 10^{-6} \text{ C} \) and \( q_2 = 1.48 \times 10^{-5} \text{ C} \).

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Key Concepts

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

Electric Charge
Electric charge is a fundamental property of matter that causes it to experience a force when placed in an electric field.
It is measured in coulombs (C) and two types exist: positive and negative.
This property is pivotal in determining how objects will interact through electric forces.
  • Like charges, both positive or both negative, will repel each other.
  • Opposite charges will attract each other, such as a positive and a negative charge.
The amount of charge on an object can vary, influencing the magnitude of forces it experiences and exerts.
For example, in the exercise, two plastic spheres with positive charges repel each other. Knowing their charges helps us understand and calculate the force experienced.
Repulsive Force
The repulsive force is observed between like charges, where two similarly charged objects push away from each other. This force can be measured and calculated using Coulomb's Law, as seen in the given exercise.
This type of force is opposite to an attractive force, which occurs between unlike charges.
  • Repulsion occurs when particles have charges of the same sign.
  • The size of the repulsive force increases with greater charge magnitudes and decreases with the square of the distance between them.
This principle is crucial in various applications, including electrostatics and electricity. Understanding the nature of these forces allows us to manipulate electronic devices and understand natural phenomena.
Point Charges
Point charges are idealized charges located at a single point in space, making the analysis of electric forces much simpler.
They are used in theoretical and instructional contexts to focus on the principles of charge and force without the complexities of real-world charge distributions.
  • They are useful for calculating forces and electric fields in controlled scenarios.
  • The concept assumes all charge is concentrated at a single point with no physical size.
In the exercise, two charged spheres are treated as point charges to apply Coulomb's Law effectively.
This allows us to easily calculate the forces between them based on their charge values and separation distance.
Coulomb's Constant
Coulomb's constant, denoted as \( k \), is a proportionality factor in Coulomb's Law.
It relates the force between two point charges to the product of their charges and the inverse square of the distance between them.
Its value is approximately \( 8.988 \times 10^9 \) Nm²/C².
  • It provides a means to calculate the strength of the electric force in the International System of Units (SI).
  • The constant simplifies the calculation of electric forces in theoretical problems and practical applications.
By understanding Coulomb's constant, students can more accurately predict and quantify electric interactions, as demonstrated in the step-by-step solution to the given exercise.

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Most popular questions from this chapter

A \(-\)3.00-nC point charge is on the \(x\)-axis at \(x =\) 1.20 m. A second point charge, \(Q,\) is on the \(x\)-axis at -0.600 m. What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) 45.0 N\(/\)C in the \(+\)x-direction, (b) 45.0 N\(/\)C in the \(-\)x-direction?

A charge of \(-\)3.00 nC is placed at the origin of an \(xy\)-coordinate system, and a charge of 2.00 nC is placed on the \(y\)-axis at \(y =\) 4.00 cm. (a) If a third charge, of 5.00 nC, is now placed at the point \(x =\) 3.00 cm, \(y =\) 4.00 cm, find the \(x\)- and \(y\)-components of the total force exerted on this charge by the other two charges. (b) Find the magnitude and direction of this force.

Two identical spheres are each attached to silk threads of length \(L =\) 0.500 m and hung from a common point (Fig. P21.62). Each sphere has mass \(m =\) 8.00 g. The radius of each sphere is very small compared to the distance between the spheres, so they may be treated as point charges. One sphere is given positive charge \(q_1\) , and the other a different positive charge \(q_2\) ; this causes the spheres to separate so that when the spheres are in equilibrium, each thread makes an angle \(\theta = 20.0^\circ\) with the vertical. (a) Draw a free-body diagram for each sphere when in equilibrium, and label all the forces that act on each sphere. (b) Determine the magnitude of the electrostatic force that acts on each sphere, and determine the tension in each thread. (c) Based on the given information, what can you say about the magnitudes of \(q_1\) and \(q_2\)? Explain. (d) A small wire is now connected between the spheres, allowing charge to be transferred from one sphere to the other until the two spheres have equal charges; the wire is then removed. Each thread now makes an angle of 30.0\(^\circ\) with the vertical. Determine the original charges. (\(Hint\): The total charge on the pair of spheres is conserved.)

Inkjet printers can be described as either continuous or drop-on-demand. In a continuous inkjet printer, letters are built up by squirting drops of ink at the paper from a rapidly moving nozzle. You are part of an engineering group working on the design of such a printer. Each ink drop will have a mass of 1.4 \(\times \space 10^{-8}\) g. The drops will leave the nozzle and travel toward the paper at 50 m\(/\)s, passing through a charging unit that gives each drop a positive charge \(q\) by removing some electrons from it. The drops will then pass between parallel deflecting plates, 2.0 cm long, where there is a uniform vertical electric field with magnitude 8.0 \(\times \space 10^4 \space N/C\). Your team is working on the design of the charging unit that places the charge on the drops. (a) If a drop is to be deflected 0.30 mm by the time it reaches the end of the deflection plates, what magnitude of charge must be given to the drop? How many electrons must be removed from the drop to give it this charge? (b) If the unit that produces the stream of drops is redesigned so that it produces drops with a speed of 25 \(m/s\), what \(q\) value is needed to achieve the same 0.30-mm deflection?

In a rectangular coordinate system a positive point charge \(q = 6.00 \times \space 10^{-9} C\) is placed at the point \(x = +0.150 m, y = 0,\) and an identical point charge is placed at \(x = -0.150 m, y = 0.\) Find the \(x\)- and \(y\)-components, the magnitude, and the direction of the electric field at the following points: (a) the origin; (b) \(x =\) 0.300 m, \(y =\) 0; (c) \(x =\) 0.150 m, \(y = -\)0.400 m; (d) \(x = 0, \)y =$ 0.200 m.

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