Problem 88 - University Physics with Modern Physics

Question

A negative point charge \(q_{1}=-4.00 \mathrm{nC}\) is on the \(x\) -axis at \(x=0.60 \mathrm{m} .\) A second point charge \(q_{2}\) is on the \(x\) -axis at \(x=-1.20 \mathrm{m} .\) What must the sign and magnitude of \(q_{2}\) be for the net electric field at the origin to be (a) 50.0 \(\mathrm{N} / \mathrm{C}\) in the \(+x\) -direction and \((\mathrm{b}) 50.0 \mathrm{N} / \mathrm{C}\) in the \(-x\) -x-direction?

Step-by-Step Solution

Answer

(a) Positive 24.0 nC; (b) Negative 8.0 nC.

Step 1: Understanding the Electric Field Contribution from q1

The electric field contribution from the charge \( q_1 = -4.00 \mathrm{nC} \) at the origin, located at \( x = 0.60 \mathrm{m} \), can be calculated using Coulomb's law. The formula is \( E = \frac{k|q|}{r^2} \), where \( k \) is the Coulomb's constant (\( k = 8.99 \times 10^9 \mathrm{N} \cdot \mathrm{m}^2/\mathrm{C}^2 \)), \( q \) is the charge magnitude, and \( r \) is the distance from the charge to the point of interest. Here, \( q_1 \) creates an electric field at the origin toward it, since it's negative.

Step 2: Calculate the Electric Field from q1 at Origin

Calculate the electric field due to \( q_1 \).\[ E_1 = \frac{k \cdot |q_1|}{(0.60)^2} \]Substituting the values:\[ E_1 = \frac{8.99 \times 10^9 \cdot 4 \times 10^{-9}}{0.36} = 99.9 \mathrm{N/C} \]The electric field direction is toward the charge \( q_1 \), thus in the \(-x\)-direction.

Step 3: Determine Requirements for q2's Electric Field Contribution in Scenario (a)

To achieve a net electric field of 50.0 \( \mathrm{N/C} \) in the \(+x\)-direction, the electric field from \( q_2 \) must overcome \( E_1 (-x\text{-direction})\) and have a \(+x\)-direction net outcome. Let \( E_2 \) be the electric field due to \( q_2 \) at the origin.Therefore, for net \(+x\) direction:\[ E_2 - 99.9 = 50.0 \E_2 = 149.9 \mathrm{N/C} \]

Step 4: Solve for q2's Magnitude in Scenario (a)

Using the formula for electric field:\[ E_2 = \frac{k |q_2|}{(1.20)^2} = 149.9 \]Substitute known values:\[ 149.9 = \frac{8.99 \times 10^9 \times |q_2|}{1.44} \|q_2| = \frac{149.9 \times 1.44}{8.99 \times 10^9} \|q_2| \approx 24.0 \mathrm{nC} \]For \( E_2 \) to be in the \(+x\)-direction, \( q_2 \) must be positive.

Step 5: Determine Requirements for q2's Electric Field Contribution in Scenario (b)

For the net electric field to be 50.0 \( \mathrm{N/C} \) in the \(-x\)-direction:\[ E_2 - 99.9 = -50.0 \E_2 = 49.9 \mathrm{N/C} \]

Step 6: Solve for q2's Magnitude in Scenario (b)

Using the same approach to find \( q_2 \):\[ E_2 = \frac{k |q_2|}{(1.20)^2} = 49.9 \]Substitute known values:\[ 49.9 = \frac{8.99 \times 10^9 \times |q_2|}{1.44} \|q_2| = \frac{49.9 \times 1.44}{8.99 \times 10^9} \|q_2| \approx 8.0 \mathrm{nC} \]For \( E_2 \) to be in the \(-x\)-direction, \( q_2 \) must be negative.

Key Concepts

Coulomb's Law

Coulomb's Law is essential for understanding how electric charges interact. It describes the force between two point charges. The formula is \[ F = \frac{k |q_1 q_2|}{r^2} \]where:- \(F\) is the force between the charges,- \(k\) is Coulomb's constant \( (8.99 \times 10^9 \, \text{N} \cdot \text{m}^2/\text{C}^2) \),- \(q_1\) and \(q_2\) are the magnitudes of the charges,- \(r\) is the distance between the charges.
In this exercise, we use Coulomb's law specifically to determine the electric fields resulting from point charges. The force changes depending on the sign of the charges:

• If both charges have the same sign, the force is repulsive, pushing the charges apart.
• If the charges have opposite signs, the force is attractive, pulling the charges together.

The law helps to understand the attraction and repulsion occurring between charges in space.

Point Charge

A point charge refers to a charge that is considered to be concentrated in a single point in space. It simplifies how we calculate electric fields and forces.
Point charges are theoretical constructs, useful for understanding electrical interactions without worrying about the actual size of the charge. In real life, this assumption might apply if:

• The distance between charges is large compared to their size.
• The spatial extent of the charge is negligible in analysis.

Here, the exercise involves two point charges \( q_1 = -4.00 \, \text{nC} \)and \( q_2 \), allowing us to apply formulas directly.

• \( q_1 \) is located at \( x = 0.60 \, \text{m} \) on the x-axis.
• \( q_2 \) is located at \( x = -1.20 \, \text{m} \) on the x-axis.

Understanding point charges lets us calculate electric fields efficiently, using theoretical point interactions, as shown in this exercise.

Electric Field Direction

The direction of an electric field indicates the force experienced by a positive test charge placed in the field. Electric fields point away from positive charges and towards negative charges.
For example, in this exercise, the electric field produced by the negative point charge \( q_1 = -4.00 \, \text{nC} \) at the origin points toward \( q_1 \). Therefore, it is in the \(-x\)-direction. Knowing the direction helps predict the behaviour of charges:

• Fields from positive charges repel other positive charges and attract negative charges.
• Fields from negative charges attract positive charges and repel other negative charges.

In scenario (a), we need the net field to point in the \(+x\)-direction, so \( q_2 \) must create a stronger opposing field. In scenario (b), the field should be \(-x\), complementing the field of \( q_1 \). Understanding field direction is crucial to solve problems involving multiple charges.

Net Electric Field

The net electric field at a point is the vector sum of the electric fields due to individual charges at that point. Each charge produces its field, and these fields combine;\( E_{\text{net}} = E_1 + E_2 + \ldots \)For this exercise:

• The electric field at the origin due to \( q_1 \) is \(99.9 \, \text{N/C}\) in the \(-x\)-direction.
• Scenario (a) requires \( E_{\text{net}} = 50.0 \, \text{N/C} \) in the \(+x\)-direction.
• Scenario (b) needs \( E_{\text{net}} = 50.0 \, \text{N/C} \) in the \(-x\)-direction.

To achieve these net fields, we adjust \( q_2 \) appropriately, considering both its magnitude and sign to counter or complement the field from \( q_1 \). Calculating the net electric field helps in understanding charge interactions and predicting the electric field configuration at any point in space.