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Point charges, \(q_{1}\) and \(q_{2}\), are placed on the \(x\) axis, with \(q_{1}\) at \(x=0\) and \(q_{2}\) at \(x=d .\) A third point charge, \(+Q\), is placed at \(x=3 d / 4 .\) If the net electrostatic force experienced by the charge \(+Q\) is zero, how are \(q_{1}\) and \(q_{2}\) related?

Short Answer

Expert verified
\(q_2 = \frac{q_1}{9}.\)

Step by step solution

01

Identify Forces Acting on +Q

Let's first note that the charge \(+Q\) is experiencing electrostatic forces due to both \(q_1\) and \(q_2\). These forces should equal in magnitude and opposite in direction to ensure the net force on \(+Q\) is zero. We'll denote the force due to \(q_1\) as \(F_{1Q}\) and due to \(q_2\) as \(F_{2Q}\).
02

Calculate Force due to q1

The force \(F_{1Q}\) acting on \(+Q\) due to \(q_1\) can be calculated using Coulomb's Law: \[ F_{1Q} = k \cdot \frac{|q_1| \cdot |Q|}{(3d/4)^2} = k \cdot \frac{|q_1| \cdot |Q|}{9d^2/16} = \frac{16k|q_1| |Q|}{9d^2}, \] where \(k\) is Coulomb's constant.
03

Calculate Force due to q2

The force \(F_{2Q}\) acting on \(+Q\) due to \(q_2\) is also given by Coulomb's Law: \[ F_{2Q} = k \cdot \frac{|q_2| \cdot |Q|}{(d/4)^2} = k \cdot \frac{|q_2| \cdot |Q|}{d^2/16} = \frac{16k |q_2| |Q|}{d^2}. \]
04

Equate Forces for Zero Net Force

For the net force on \(+Q\) to be zero, the magnitudes of \(F_{1Q}\) and \(F_{2Q}\) should be equal: \[ \frac{16k |q_1| |Q|}{9d^2} = \frac{16k |q_2| |Q|}{d^2}. \] Simplifying this equation by canceling common factors gives us: \[ \frac{|q_1|}{9} = |q_2|. \] So, \(|q_2| = \frac{|q_1|}{9}.\)
05

Provide the Relation between q1 and q2

From the equation \(|q_2| = \frac{|q_1|}{9}\), it follows that the magnitude of \(q_2\) is one-ninth of the magnitude of \(q_1\). Therefore, the relationship between the charges is: \[ q_2 = \frac{q_1}{9}. \]

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Key Concepts

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

Coulomb's Law
Coulomb's Law is a fundamental principle in electrostatics. It describes the electrostatic force between two point charges. The law states that the magnitude of the force, often noted as \(F\), between two charges \(q_1\) and \(q_2\) is proportional to the product of the absolute magnitudes of the charges, and inversely proportional to the square of the distance \(r\) between them:
\[ F = k \frac{|q_1| \, |q_2|}{r^2}, \]
where \(k\) is Coulomb's constant, equal to \(8.9875 \times 10^9 \, \text{N m}^2/\text{C}^2\).

  • This law helps us calculate forces in systems with multiple charges, like the one in the exercise, where more than two charges interact with each other.
  • It applies to point charges, which are idealized charges located at a single point in space, simplifying complex systems for easier analysis.
  • Coulomb's Law forms the basis for understanding interactions in electric fields.
Point Charge
A point charge is a charge located at a single point in space. It simplifies complex electric field calculations by assuming the charge size is negligible compared to the distances involved. In practical terms, whenever the dimensions of the charged object are very small compared to the distances between objects, it can often be modeled as a point charge.

In the exercise, \(q_1\), \(q_2\), and \(+Q\) are all treated as point charges to easily apply Coulomb's Law. The simplicity of a point charge is useful because:

  • It allows for straightforward application of mathematical formulas like Coulomb's Law.
  • It is ideal in theoretical studies and problem-solving in physics, providing clear and precise behaviors for electric forces.
By treating each charge as a point, calculations become more manageable, ensuring accuracy when predicting the behavior of electrostatic forces.
Electrostatic Force
Electrostatic force is the force that interacts between stationary charged bodies. Its nature is either attractive or repulsive, depending on the polarity of the charges involved—like charges repel each other and opposite charges attract.
The importance of electrostatic force in electrostatics lies in its influence on how charges move and distribute themselves within electric fields.

Key Characteristics:

  • It follows an inverse square law (as seen in Coulomb's Law), where forces decrease with the square of the distance from the source charge.
  • Electrostatic force is central to understanding phenomena in electric circuits, chemistry, and fields like material science.
In the exercise, we calculate the electrostatic forces \(F_{1Q}\) and \(F_{2Q}\) acting on the charge \(+Q\). By ensuring these forces balance each other out (one must equal the other in magnitude but act in opposite directions), we determine that the net force on \(+Q\) is zero. This fundamental understanding helps students analyze similar scenarios in more complex arrangements.

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Most popular questions from this chapter

The electric field at the point \(x=5.00 \mathrm{cm}\) and \(y=0\) points in the positive \(x\) direction with a magnitude of \(10.0 \mathrm{N} / \mathrm{C} .\) At the point \(x=10.0 \mathrm{cm}\) and \(y=0\) the electric field points in the positive \(x\) direction with a magnitude of \(15.0 \mathrm{N} / \mathrm{C}\). Assuming this electric field is produced by a single point charge, find (a) its location and (b) the sign and magnitude of its charge.

An electron and a proton are released from rest in space, far from any other objects. The particles move toward each other, due to their mutual electrical attraction. (a) When they meet, is the kinetic energy of the electron greater than, less than, or equal to the kinetic energy of the proton? (b) Choose the best explanation from among the following: I. The proton has the greater mass. since kinetic energy is proportional to mass, it follows that the proton will have the greater kinetic energy. II. The two particles experience the same force, but the light electron moves farther than the massive proton. Therefore, the work done on the electron, and hence its kinetic energy, is greater. III. The same force acts on the two particles, Therefore, they will have the same kinetic energy and energy will be conserved.

An object with a charge of \(-3.6 \mu \mathrm{C}\) and a mass of \(0.012 \mathrm{kg}\) experiences an upward electric force, due to a uniform electric field, equal in magnitude to its weight. (a) Find the direction and magnitude of the electric field. (b) If the electric charge on the object is doubled while its mass remains the same, find the direction and magnitude of its acceleration.

When a nerve impulse propagates along a nerve cell, the electric field within the cell membrane changes from \(7.0 \times 10^{5} \mathrm{N} / \mathrm{C}\) in one direction to \(3.0 \times 10^{5} \mathrm{N} / \mathrm{C}\) in the other direction. Approximating the cell membrane as a parallel-plate capacitor, find the magnitude of the change in charge density on the walls of the cell membrane.

An electrically neutral object is given a positive charge. (a) In principle, does the object's mass increase, decrease, or stay the same as a result of being charged? (b) Choose the best explanation from among the following: I. To give the object a positive charge we must remove some of its electrons; this will reduce its mass. II. Since electric charges have mass, giving the object a positive charge will increase its mass. III. Charge is conserved, and therefore the mass of the object will remain the same.

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