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\(\bullet$$\bullet \mathrm{A}-5.00 \mathrm{nC}\) point charge is on the \(x\) axis at \(x=1.20 \mathrm{m} . \mathrm{A}\) second point charge \(Q\) is on the \(x\) axis at \(-0.600 \mathrm{m} .\) What must be the sign and magnitude of \(Q\) for the resultant electric field at the origin to be (a) 45.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) direction, \((\mathrm{b}) 45.0 \mathrm{N} / \mathrm{C}\) in the \(-x\) direction?

Short Answer

Expert verified
(a) \( Q \approx +3.05 \text{ nC} \); (b) \( Q \approx -0.55 \text{ nC} \).

Step by step solution

01

Understanding the Problem

We have two point charges, one at \( x = 1.20 \text{ m} \) with charge \( A = -5.00 \text{ nC} \) and a second charge \( Q \) at \( x = -0.60 \text{ m} \). We need to find the sign and magnitude of \( Q \) such that the resultant electric field at the origin is 45.0 N/C in the given directions.
02

Electric Field Due to a Point Charge

The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is given by \( E = \frac{k|q|}{r^2} \), where \( k = 8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \) is the Coulomb's constant. The direction of \( E \) is away from the charge for a positive charge and towards the charge for a negative charge.
03

Calculate the Electric Field from Charge A

For charge \( A \) at \( x = 1.20 \text{ m} \), the electric field at the origin \( E_A \) is towards the charge (since \( A \) is negative). Thus, using the formula: \[ E_A = \frac{k |-5.00 \times 10^{-9} \text{ C}|}{(1.20)^2} \approx 31.22 \text{ N/C towards } \] the negative direction (or \(-x\)).
04

Set up the Equation for the Resultant Electric Field - Part (a)

For part (a), we want the total electric field \( E_{total} = 45.0 \text{ N/C} \) in the \(+x\) direction. Therefore, the electric field due to \( Q \), \( E_Q \), must be such that:\[ E_Q - E_A = 45.0 \]Substitute \( E_A \) from the previous step:\[ E_Q - 31.22 = 45.0 \] This gives \( E_Q = 76.22 \text{ N/C} \) pointing towards \(+x\).
05

Calculate Magnitude and Sign of Q for Part (a)

Using \( E_Q = 76.22 \text{ N/C} \) and the distance \( r = 0.60 \text{ m} \), from the origin to charge \( Q \), substitute into the formula:\[ E_Q = \frac{k |Q|}{(0.60)^2} = 76.22 \]Solving for \( |Q| \):\[ |Q| = \frac{76.22 \times 0.36}{8.99 \times 10^9} \approx 3.05 \text{ nC} \]Since the field is in the positive direction, charge \( Q \) must be positive.
06

Set up the Equation for the Resultant Electric Field - Part (b)

For part (b), \( E_{total} = 45.0 \text{ N/C} \) in the \(-x\) direction. Thus:\[ E_A - E_Q = 45.0 \]Substitute \( E_A \):\[ 31.22 - E_Q = 45.0 \] This gives \( E_Q = -13.78 \text{ N/C} \), pointing towards \(-x\).
07

Calculate Magnitude and Sign of Q for Part (b)

With \( E_Q = 13.78 \text{ N/C} \) and the same distance, use the formula:\[ -E_Q = \frac{k |Q|}{(0.60)^2} = 13.78 \]Recalculate magnitude:\[ |Q| = \frac{13.78 \times 0.36}{8.99 \times 10^9} \approx 0.55 \text{ nC} \]Since \( E_Q \) must point towards \(-x\), \( Q \) must be negative.

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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 that helps us understand the electric forces between charges. It states that the electric force between two point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them. The formula is represented as:\[F = k \frac{|q_1 q_2|}{r^2}\]where
  • \( F \) is the force between the charges.
  • \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \, \text{Nm}^2/\text{C}^2 \).
  • \( q_1 \) and \( q_2 \) are the amounts of the charges.
  • \( r \) is the distance between the centers of the two charges.
This law is crucial when calculating the forces or fields generated by point charges in exercises. By understanding this fundamental concept, students can predict how charges will interact under various conditions and configurations.
Point Charge
A point charge is an idealized model of a charge that is considered to exist at a single point in space. It's a useful simplification for calculating electric fields and forces, especially when the size of the object holding the charge is negligible compared to the distance between charges. In practical scenarios, like the exercise we're covering, point charges are used to simplify the understanding of electric interactions. They allow us to apply formulas like those from Coulomb's law, assuming all the charge is concentrated at that point.
  • Point charges can be positive or negative, affecting the direction of the resultant electric field.
  • They are essential for building an understanding of electric fields and forces in both theoretical and applied electricity concepts.
Overall, considering charges as point charges aids in focusing on calculations regarding their interactions without complications of distribution over a physical object.
Electric Field Direction
The direction of an electric field is defined as the direction a positive test charge would move when placed in the field. For a positive point charge, the electric field radiates outward, while for a negative point charge, it points inward towards the charge. The significance of electric field direction becomes clear when determining the effect of multiple charges. In problems like the one we have:
  • The field direction from each charge must be considered to derive the net field at a point, such as the origin.
  • Positive charges contribute fields that push away, whereas negative charges pull towards them.
  • Understanding these directions assists in correctly determining if charges will add or subtract from the total electric field in any given direction.
Grasping these concepts helps in predicting the overall impact of multiple point charges, an essential skill in both academic and practical applications of electromagnetism.
Calculation of Electric Fields
Calculating electric fields involves using the principle that electric field magnitude due to a point charge is defined by:\[E = \frac{k|q|}{r^2}\]Where
  • \( E \) is the electric field.
  • \( k \) is Coulomb’s constant.
  • \( q \) is the point charge.
  • \( r \) is the distance from the charge to the point where the field is being calculated.
In exercises, this means determining the contribution of each individual charge to the field at a particular point and combining those effects, taking direction into account, to find a result.By analyzing such problems:
  • We hone the skill of setting up and solving equations where multiple charges affect the field directionality and magnitude.
  • We calculate the resultant field by summing the fields produced by each point charge, attending carefully to their vector nature (direction and magnitude).
This process is foundational for deeper exploration into electric fields, paving the way for more complex concepts like field lines and electromagnetism.

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Most popular questions from this chapter

\(\bullet\) Two iron spheres contain excess charge, one positive and the other negative. (a) Show how the charges are arranged on these spheres if they are very far from each other. (b) If the spheres are now brought close to each other, but do not touch, on the copper ball. sketch how the charges will be distributed on their surfaces. (c) In part (b), show how the charges would be distributed if both spheres were negative.

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