Problem 39 - University Physics with Modern Physics

Question

A \(+\)2.00-nC point charge is at the origin, and a second \(-\)5.00-nC point charge is on the \(x\)-axis at \(x = \)0.800 m. (a) Find the electric field (magnitude and direction) at each of the following points on the x-axis: (i) \(x =\) 0.200 m; (ii) \(x =\) 1.20 m; (iii) \(x = -\)0.200 m. (b) Find the net electric force that the two charges would exert on an electron placed at each point in part (a).

Step-by-Step Solution

Answer

(i) 574.36 N/C right, 9.19x10^-17 N left; (ii) 268.45 N/C left, 4.30x10^-17 N right; (iii) 404.55 N/C left, 6.47x10^-17 N right.

Step 1: Calculating the Electric Field by a Point Charge

The electric field \( \vec{E} \) created by a point charge \( q \) at a distance \( r \) is given by the formula: \[ E = \frac{k \cdot |q|}{r^2} \] where \( k = 8.99 \times 10^9 \ \text{N} \cdot \text{m}^2/\text{C}^2 \) is the Coulomb's constant. It is important that the direction of \( \vec{E} \) is radial, pointing away from \( q \) if \( q \) is positive and towards \( q \) if it is negative.

Step 2: Electric Field at x = 0.200 m

Consider position \( x = 0.200 \ m \):- For the +2.00 nC charge: - Distance \( r_1 = 0.200 \ m \) - \( E_1 = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-9}}{0.200^2} = 449.5 \ \text{N/C} \), directed away towards positive x-axis.- For the -5.00 nC charge: - Distance \( r_2 = 0.800 - 0.200 = 0.600 \ m \) - \( E_2 = \frac{8.99 \times 10^9 \times 5.00 \times 10^{-9}}{0.600^2} = 124.86 \ \text{N/C} \), directed towards the negative charge, towards positive x-axis.Net \( E = E_1 + E_2 = 449.5 + 124.86 = 574.36 \ \text{N/C} \) towards the positive x-axis.

Step 3: Electric Field at x = 1.20 m

Consider position \( x = 1.20 \ m \):- For the +2.00 nC charge: - Distance \( r_1 = 1.20 \ m \) - \( E_1 = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-9}}{1.20^2} = 12.49 \ \text{N/C} \), directed towards positive x-axis.- For the -5.00 nC charge: - Distance \( r_2 = 1.20 - 0.800 = 0.400 \ m \) - \( E_2 = \frac{8.99 \times 10^9 \times 5.00 \times 10^{-9}}{0.400^2} = 280.94 \ \text{N/C} \), directed towards the charge, hence towards negative x-axis.Net \( E = E_2 - E_1 = 280.94 - 12.49 = 268.45 \ \text{N/C} \), directed towards the negative x-axis.

Step 4: Electric Field at x = -0.200 m

Consider position \( x = -0.200 \ m \):- For the +2.00 nC charge: - Distance \( r_1 = 0.200 \ m \) - \( E_1 = \frac{8.99 \times 10^9 \times 2.00 \times 10^{-9}}{0.200^2} = 449.5 \ \text{N/C} \), directed towards negative x-axis.- For the -5.00 nC charge: - Distance \( r_2 = 0.800 - (-0.200) = 1.000 \ m \) - \( E_2 = \frac{8.99 \times 10^9 \times 5.00 \times 10^{-9}}{1.000^2} = 44.95 \ \text{N/C} \), directed towards positive x-axis.Net \( E = E_1 - E_2 = 449.5 - 44.95 = 404.55 \ \text{N/C} \), directed towards negative x-axis.

Step 5: Electric Force on an Electron

The force \( \vec{F} \) on an electron in an electric field \( \vec{E} \) is given by \( \vec{F} = e \cdot \vec{E} \), where \( e = 1.60 \times 10^{-19} \ \text{C} \) is the elementary charge. Calculate for each point:(i) At \( x = 0.200 \ m \):\( F = 1.60 \times 10^{-19} \times 574.36 = 9.19 \times 10^{-17} \ \text{N} \) towards negative x-axis.(ii) At \( x = 1.20 \ m \):\( F = 1.60 \times 10^{-19} \times 268.45 = 4.295 \times 10^{-17} \ \text{N} \) towards positive x-axis.(iii) At \( x = -0.200 \ m \):\( F = 1.60 \times 10^{-19} \times 404.55 = 6.473 \times 10^{-17} \ \text{N} \) towards positive x-axis.

Key Concepts

Coulomb's law

Coulomb's law helps us understand the electric interaction between charged particles. The formula is given by: \[F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\]where:

• \( F \) is the electric force between two charges.
• \( k = 8.99 \times 10^9 \ \text{N} \cdot \text{m}^2/\text{C}^2 \) is Coulomb's constant, a value that quantifies the electric force per unit charge per distance squared.
• \( q_1 \) and \( q_2 \) are the magnitudes of the charges.
• \( r \) is the separation between the charges.

The electric force is directly proportional to the product of the charges and inversely proportional to the square of their distance. This law allows us to calculate the strength and direction of forces acting between charged particles. Just remember that like charges repel while opposite charges attract. This fundamental principle explains a lot about interactions in the world of charged particles.
The equation also highlights how distance affects the force. Double the distance, and you'll see the force reduce to a quarter of its original value.

Electric force

Electric force is a key concept in electricity and magnetism. It is the force exerted by charged particles on one another. When two charges interact, the force they exert can be calculated using Coulomb’s law.
The direction of this force depends on the types of charges involved. If both charges are positive, they will push each other away (repulsion). If one charge is positive and the other negative, they will pull each other closer (attraction).
The magnitude of electric force is influenced by two main factors: the sizes of the charges and the distance between them.

• The greater the absolute value of each charge, the stronger the force between them.
• The closer the charges, the stronger the force, since it varies inversely with the square of the distance.

In practical applications, understanding electric force helps in designing circuits, capacitors, and other electrical systems. It is crucial to ensure that electrical components are correctly positioned to avoid undesired repulsion or attraction in devices, especially when dealing with charged objects.

Point charge

A point charge is an idealized model of a charged particle. It assumes that the charge is concentrated at a single point in space. This simplification is very useful for theoretical calculations, allowing us to examine the effects of electric fields and forces without worrying about the physical size of the charge itself.
When calculating the electric field created by a point charge, we use the formula:\[E = \frac{k \cdot |q|}{r^2} \]where:

• \( E \) is the electric field intensity due to the point charge.
• \( q \) is the magnitude of the point charge.
• \( r \) is the distance from the point charge to the point where the field is being measured.

The electric field is a vector quantity, which means it has both magnitude and direction. Around a positive point charge, the field radiates outward, while around a negative charge, the field converges inward.
Understanding how point charges work simplifies complex problems in electrostatics and helps us solve practical physics problems like finding the resultant field or force at any given point in the field around multiple charges.