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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?

Short Answer

Expert verified
(a) +2.55 nC (b) -1.05 nC

Step by step solution

01

Understanding the Problem

We have two point charges, one fixed at 1.20 m on the positive x-axis with a charge of \(-3.00\,\text{nC}\), and another charge \(Q\) is on the negative x-axis at \(-0.600\,\text{m}\). We need to find the magnitude and sign of \(Q\) so that the net electric field at the origin is \(45.0\,\text{N/C}\) either in the \(+x\)-direction or \(-x\)-direction.
02

Formula for Electric Field due to a Point Charge

The electric field \( E \) due to a point charge \( q \) at a distance \( r \) is given by the formula: \( E = \frac{k \cdot |q|}{r^2} \), where \( k = 8.99 \times 10^9 \,\text{N}\cdot m^2/\text{C}^2 \) is the electrostatic constant.
03

Calculate Electric Field due to First Charge

For the charge \(-3.00\,\text{nC}\) at \(1.20\,\text{m}\), the electric field at the origin points towards the charge, hence in the \(-x\)-direction. Calculate its magnitude: \[ E_1 = \frac{8.99 \times 10^9 \times 3.00 \times 10^{-9}}{(1.20)^2} = 18.73 \, \text{N/C} \] directed along \(-x\)-axis.
04

Set Up Equation for Part (a)

For the net electric field to be \(45.0\,\text{N/C}\) in the \(+x\)-direction, the electric field due to \(Q\) must counteract the field by the first charge and add extra field strength in the \(+x\) direction. Equation: \[ E_Q - 18.73 = 45 \, \text{N/C} \] Thus,\[ E_Q = 63.73 \, \text{N/C} \]
05

Solve for Charge Q in Part (a)

Using the formula for electric field, solve for \(Q\): \[ E_Q = \frac{k\cdot |Q|}{0.600^2} \Rightarrow |Q| = E_Q \cdot \frac{(0.600)^2}{k} \]\[ |Q| = 63.73 \times \frac{0.360}{8.99 \times 10^9} = 2.55 \times 10^{-9} \, \text{C} = 2.55\,\text{nC} \]Because the field needs to be in the \(+x\)-direction, \(Q\) must be positive.
06

Set Up Equation for Part (b)

For the resultant electric field to be \(45.0 \, \text{N/C}\) in the \(-x\)-direction, the contribution from \(Q\) must still be less than the electric field due to the first charge, amplifying its effect.Equation: \[ -E_Q - 18.73 = -45 \, \text{N/C} \] Thus,\[ -E_Q = 26.27 \, \text{N/C} \]
07

Solve for Charge Q in Part (b)

Solve for \(Q\) in part (b) with the electric field equation:\[ E_Q = 26.27 \Rightarrow |Q| = E_Q \cdot \frac{(0.600)^2}{k} \]\[ |Q| = 26.27 \times \frac{0.360}{8.99 \times 10^9} = 1.05 \times 10^{-9} \, \text{C} = 1.05\,\text{nC} \]Since the field should remain in the \(-x\)-direction, \(Q\) must be negative.

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Key Concepts

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

Electric Field
The electric field is a fundamental concept in electrostatics. It is a vector field that shows how a point charge or distribution of charges affects the surrounding space. Think of it as the region around a charge where its electric force can be felt by other charges.

The electric field produced by an isolated charge can be calculated using the expression:
  • \( E = \frac{k \cdot |q|}{r^2} \)
  • where \(E\) is the electric field, \(k\) is Coulomb's constant, \(|q|\) is the magnitude of the charge, and \(r\) is the distance from the charge.
For any point charge, the direction of the electric field is along the line connecting the charge and the point where the field is being calculated. If the charge is positive, the field points away; if negative, it points towards the charge. At the origin of this problem, our goal was to find how another charge, \(Q\), can ensure that the resultant electric field meets specific criteria.
Point Charge
In physics, a point charge refers to a charged object with negligible size. This idealization helps in solving electrostatic problems with simplicity and precision.

In the context of this exercise, point charges are used to model the individual charges located on the x-axis.

They are:
  • The first point charge of \(-3.00 \,\text{nC}\) is placed at \(x = 1.20 \,\text{m}\).
  • The second charge, \(Q\), whose sign and magnitude are the subjects of this exercise, is placed at \(x = -0.600 \,\text{m}\).
When dealing with such charges, the interaction and resulting fields can be effectively predicted using simple algebraic calculations and known laws of electrostatics, like Coulomb’s Law.
Coulomb's Law
Coulomb's Law is central to understanding electric forces and fields. It calculates the force between two stationary charges. The law states that the force \( F \) between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. Mathematically, it is expressed as:
  • \( F = k \cdot \frac{|q_1 \cdot q_2|}{r^2} \)
  • where \(F\) is the force, \(q_1\) and \(q_2\) are the amounts of the charges, \(r\) is the distance between the charges, and \(k\) is Coulomb's constant, which has a value of \(8.99 \times 10^9 \,\text{N} \cdot \text{m}^2/\text{C}^2\).
This law allows us to determine the magnitude and direction of the electric force between the two charges, which is what guides the calculation of the electric field at the origin in this exercise.
Electric Forces
Electric forces explain the interaction between charged objects. When dealing with static (non-moving) charges, these forces are a consequence of electric fields. They can either be attractive or repulsive:
  • Attractive forces occur between opposite charges (positive and negative).
  • Repulsive forces happen between like charges (both positive or both negative).
In this exercise, the electric force is significant in determining the overall effect of the two charges at the origin. The aim was to modify the second charge, \(Q\), so the combined forces result in a specific electric field direction and magnitude. By carefully balancing the magnitudes and directions of the forces between the two charges and the point at the origin, we can predictably control field outcomes using these fundamental electrostatic principles.

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Most popular questions from this chapter

A proton is placed in a uniform electric field of 2.75 \(\times 10^3 \space N/C\). Calculate (a) the magnitude of the electric force felt by the proton; (b) the proton's acceleration; (c) the proton's speed after 1.00 \(\mu\)s in the field, assuming it starts from rest.

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.

Three identical point charges \(q\) are placed at each of three corners of a square of side \(L\). Find the magnitude and direction of the net force on a point charge \(-3q\) placed (a) at the center of the square and (b) at the vacant corner of the square. In each case, draw a free-body diagram showing the forces exerted on the \(-3q\) charge by each of the other three charges.

Two small spheres with mass \(m =\) 15.0 g are hung by silk threads of length \(L =\) 1.20 m from a common point (Fig. P21.62). When the spheres are given equal quantities of negative charge, so that \(q_1 = q_2 = q\), each thread hangs at \(\theta = 25.0^\circ\) from the vertical. (a) Draw a diagram showing the forces on each sphere. Treat the spheres as point charges. (b) Find the magnitude of \(q\). (c) Both threads are now shortened to length \(L =\) 0.600 m, while the charges \(q_1\) and \(q_2\) remain unchanged. What new angle will each thread make with the vertical? (\(Hint\): This part of the problem can be solved numerically by using trial values for \(\theta\) and adjusting the values of \(\theta\) until a self-consistent answer is obtained.)

If a proton and an electron are released when they are 2.0 x 10\(^{-10}\) m apart (a typical atomic distance), find the initial acceleration of each particle.

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