Problem 81 - University Physics with Modern Physics

Question

A negative point charge \(q_1 = -4.00\) nC is on the \(x\)-axis at \(x =\) 0.60 m. A second point charge \(q_2\) is on the \(x\)-axis at \(x = -\)1.20 m. What must the sign and magnitude of \(q_2\) be for the net electric field at the origin to be (a) 50.0 N\(/\)C in the \(+x\)-direction and (b) 50.0 N\(/\)C in the \(-\)x-direction?

Step-by-Step Solution

Answer

For (a), \(q_2 = -9.6\) nC. For (b), \(q_2 = +9.6\) nC.

Step 1: Understanding the Problem

We need to find the charge \(q_2\) such that the net electric field at the origin is a specific value in the given directions. Electric fields due to point charges are influenced by both the magnitude and sign of the charge. A negative charge generates an electric field pointing towards it, while a positive charge generates an electric field pointing away from it.

Step 2: Calculate Electric Field from Charge q1

The electric field \(E_1\) from charge \(q_1\) at the origin can be calculated using the formula: \[ E = \frac{k|q|}{r^2} \]where \(k = 8.99 \times 10^9\) N·m²/C², \(q_1 = -4.00 \) nC, and \(r = 0.60\) m.Conduct calculations:\[ E_1 = \frac{8.99 \times 10^9 \times 4.00 \times 10^{-9}}{(0.60)^2} \approx 9.99 \, \text{N/C} \]Since \(q_1\) is negative, the field points towards \(q_1\), thus in the \(-x\) direction at the origin.

Step 3: Set Up Net Electric Field Equation for Part (a)

For the net electric field to be 50.0 N/C in the \(+x\) direction, the field due to \(q_2\) must overcome the field due to \(q_1\) and point towards the \(+x\) direction. Therefore, \[ E_1 - E_2 = -50.0 \, \text{N/C} \]Solving for \(E_2\):\[E_2 = E_1 + 50.0 = 9.99 + 50.0 = 59.99 \, \text{N/C}\].

Step 4: Calculate Required q2 for Part (a)

Using the electric field formula for \(E_2\) and solving for \(q_2\), we assume \(r = 1.20\) m for \(x = -1.20\) m:\[ \frac{k|q_2|}{(1.20)^2} = 59.99 \, \text{N/C} \]Substitute values:\[ |q_2| = \frac{59.99 \times (1.20)^2}{8.99 \times 10^9} \]Calculate:\[ |q_2| = 9.6 \, \text{nC} \]The charge \(q_2\) must be negative to create an electric field towards the \(-x\) direction at the origin.

Step 5: Verify Part (b) Setup and Requirements

For the net electric field to be 50.0 N/C in the \(-x\) direction, the charge \(q_2\) should be such that:\[E_2 - E_1 = 50.0 \, \text{N/C}\]Thus:\[E_2 = 50.0 + 9.99 = 59.99 \, \text{N/C}\].

Step 6: Calculate Required q2 for Part (b)

Using the electric field formula for \(E_2\) again:\[ \frac{k|q_2|}{(1.20)^2} = 59.99 \, \text{N/C} \]Substitute values as before:\[ |q_2| = 9.6 \, \text{nC} \]Here, the charge \(q_2\) must be positive to ensure the electric field directs away, i.e., towards the \(+x\) direction at the origin.

Key Concepts

Point Charge

Point charges are like small bundles of electric charge concentrated at a single point in space. They are fundamental in understanding electric phenomena because they can approximately model many scenarios involving charged particles like electrons and protons.

A point charge creates an electric field around it, influencing other charges or charged objects nearby. The magnitude of this field depends on the amount of charge and the distance from the charge. For example, a charge of each point charge, denoted as \( q \), could be either positive or negative.

Positive charges traditionally produce an electric field that points outward, while negative charges create an electric field that points inward towards the charge. Knowing the nature and position of point charges can help to determine how they interact with each other and influence their surroundings.

Electric Field Direction

The direction of an electric field is a fundamental concept that describes the flow of electric potential around charge distributions. Understanding how electric fields point helps us to predict how forces act on charges within the field.

Electric fields are vector fields represented by arrows showing direction and magnitude. The direction is defined as the path that a positive test charge would take under the influence of the field. Therefore, for a positive point charge, the electric field direction extends outward. Conversely, it points inward towards a negative point charge.

In solving physics problems, like the one given, examining the direction of electric fields helps to determine the resulting net electric field at a particular location. If you have multiple charges affecting a point, the net electric field direction is the vector sum of all individual fields at that point.

Coulomb's Law

Coulomb's Law is a principle that quantifies the electric force between two point charges. It's essential for calculating how charged objects interact at a distance. The law states that the electric force ( \( F \) ) between two point charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. This can be represented mathematically as:

\[ F = \frac{k |q_1 q_2|}{r^2} \]

where:

• \( k \) is Coulomb's constant, approximately \( 8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2 \).
• \( q_1 \) and \( q_2 \) are the two point charges.
• \( r \) is the distance between the charges.

Applying Coulomb's Law allows us to calculate the electric fields generated by point charges, as in the problem at hand. Using this calculation, you can also determine the net effect of multiple charges on a point in space by considering each charge's contribution to the total electric field.