Problem 68 - University Physics with Modern Physics

Question

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?

Step-by-Step Solution

Answer

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

Step 1: 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.

Step 2: 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.

Step 3: 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.

Step 4: 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,\[ E_Q = 63.73 \, \text{N/C} \]

Step 5: 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.

Step 6: 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,\[ -E_Q = 26.27 \, \text{N/C} \]

Step 7: 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.

Key Concepts

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.