Problem 72 - University Physics with Modern Physics

Question

\(\mathrm{A}-5.00\) -nC point charge is on the \(x\) -axis at \(x=1.20 \mathrm{m.}\) 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 \(+x\) -x-direction, (b) 45.0 \(\mathrm{N} / \mathrm{C}\) in the - \(x\) -direction?

Step-by-Step Solution

Answer

(a) Q is -3.06 nC. (b) Q is -0.555 nC.

Step 1: Understand the Electric Field Equation

The electric field created by a point charge at a distance is given by the equation \( E = \frac{k |q|}{r^2} \), where \( k = 8.99 \times 10^9\, \mathrm{N}\, \mathrm{m}^2/\mathrm{C}^2 \) is Coulomb's constant, \( q \) is the charge, and \( r \) is the distance from the charge.

Step 2: Calculate the Electric Field due to A at the Origin

We have a charge \( A = -5.00\, \mathrm{nC} \) located at \( x = 1.20\, \mathrm{m} \). The distance from \( A \) to the origin is \( 1.20\, \mathrm{m} \). The electric field due to charge \( A \) is:\[ E_A = \frac{k \cdot 5.00 \times 10^{-9}\, \mathrm{C}}{(1.20)^2} = \frac{8.99 \times 10^9 \cdot 5.00 \times 10^{-9}}{1.44} \approx 31.2 \ \mathrm{N/C} \]This field points towards the negative charge (the negative \( x \)-direction).

Step 3: Set the Condition for the Resultant Electric Field

To solve part (a), we need the resultant electric field at the origin to be \( 45.0 \ \mathrm{N/C} \) in the \(+x\)-direction. Therefore, \( E_A \) and the field from \( Q \), \( E_Q \), combine to give a net electric field of \( 45.0 \ \mathrm{N/C} \) to the right.

Step 4: Calculate Required Electric Field by Q, Part (a)

For the net field to be \( 45.0 \ \mathrm{N/C} \) in the \(+x\)-direction, \( E_Q \) must counteract \( E_A \) and exceed it in the \(+x\)-direction:\[ E_Q - 31.2 = 45.0 \Rightarrow E_Q = 76.2 \ \mathrm{N/C} \]

Step 5: Determine the Magnitude and Sign of Q, Part (a)

The distance of \( Q \) from the origin is \( 0.600\, \mathrm{m} \). Using \( E_Q = \frac{k |Q|}{(0.600)^2} \), we solve for \( |Q| \):\[ |Q| = \frac{76.2 \cdot (0.600)^2}{8.99 \times 10^9} = 3.06 \times 10^{-9} \ \mathrm{C} \approx 3.06 \ \mathrm{nC} \]Since \( E_Q \) must point in the \(+x\)-direction to add to \( 45.0 \ \mathrm{N/C} \), \( Q \) must be negative (point towards the origin).

Step 6: Calculate Electric Field for Part (b)

For part (b), the resultant electric field at the origin should be \(-45.0 \ \mathrm{N/C} \). Therefore, \( E_Q \) must support \( E_A \) in the \(-x\)-direction:\[ -E_A - E_Q = -45.0 \Rightarrow E_Q = 13.8 \ \mathrm{N/C} \]

Step 7: Determine the Magnitude and Sign of Q, Part (b)

Again, using \( E_Q = \frac{k |Q|}{(0.600)^2} \), solve for \( |Q| \):\[ |Q| = \frac{13.8 \cdot (0.600)^2}{8.99 \times 10^9} = 0.555 \times 10^{-9} \ \mathrm{C} \approx 0.555 \ \mathrm{nC} \]This field must point in the \(-x\)-direction, so \( Q \) should be negative, reinforcing \( E_A \).

Key Concepts

Coulomb's law

Coulomb's law is a fundamental principle in electromagnetism that describes how electric charges interact. Named after the French physicist Charles-Augustin de Coulomb, this law quantifies the amount of force between two charges. The force (\(F\)) experienced by two point charges is directly proportional to the product of their magnitudes (\(q_1\) and \(q_2\)) and inversely proportional to the square of the distance (\(r^2\)) between them. Mathematically, it is expressed as:

• \[F = k \frac{|q_1 q_2|}{r^2}\]

where \(k = 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2\) is the Coulomb's constant. This formula shows that as the charge values increase, the force becomes stronger, whereas a greater distance weakens the force. This law is a cornerstone in understanding how electric fields arise and interact in different scenarios.

point charge

A point charge is an idealized model of a particle with an electric charge concentrated at a single point in space. This simplification helps in calculating electric fields and forces without the complexity of considering the charge's physical dimensions. In reality, no charge is infinite small, but at large distances, many charged objects behave like point charges.A point charge generates an electric field (\(E\)), which describes how the charge influences other charges around it. This is given by the equations we use from Coulomb's law. The electric field created by a point charge diminishes as one moves farther away from the charge, following the inverse square law as seen in:

• \[E = \frac{k |q|}{r^2}\]

In practical terms, when we consider the behavior of a point charge, it helps in understanding important concepts such as the direction and magnitude of electric fields and how they combine when multiple charges are present, like in the original exercise.

vector addition

Vector addition is a method used to combine vectors, which have both magnitude and direction. This method is particularly useful in physics for combining forces or fields, like electric fields from multiple charges.Suppose you have two electric fields, one from charge \(A\) and the other from charge \(Q\). The total or net electric field at a point is determined through vector addition of these individual fields. Since electric field vectors can have positive or negative directions along the axis (depending on the charge's sign and location), it's crucial to add them taking into account their directions:

• If both fields face the same direction, their magnitudes add up.
• If they face opposite directions, subtract the smaller from the larger magnitude and take the direction of the larger field.

In the example provided, finding the resultant field at the origin involved adjusting the magnitude and sign of charge \(Q\) to achieve the desired field strength and direction, demonstrating the principle of vector addition effectively.