Problem 23 - College Physics

Question

A point charge \(Q=+4.60 \mu \mathrm{C}\) is held fixed at the origin. A second point charge \(q=+1.20 \mu \mathrm{C}\) with mass of \(2.80 \times\) \(10^{-4} \mathrm{kg}\) is placed on the \(x\) axis, 0.250 \(\mathrm{m}\) from the origin. (a) What is the electric potential energy \(U\) of the pair of charges? (Take \(U\) to be zero when the charges have infinite separation.) (b) The second point charge is released from rest. What is its speed when its distance from the origin is (i) \(0.500 \mathrm{m} ;\) (ii) 5.00 \(\mathrm{m}\) ; (iii) 50.0 \(\mathrm{m} ?\)

Step-by-Step Solution

Answer

(a) 0.198 J, (b) (i) 26.6 m/s, (ii) 57.8 m/s, (iii) 59.7 m/s.

Step 1: Determine Initial Electric Potential Energy

The electric potential energy \( U \) between two point charges is given by the formula \( U = \frac{kQq}{r} \), where \( k = 8.99 \times 10^9 \ \text{Nm}^2/\text{C}^2 \) is Coulomb's constant, \( Q \) and \( q \) are the point charges, and \( r \) is the distance between them. Substituting the provided values gives \( U = \frac{(8.99 \times 10^9) \times (4.60 \times 10^{-6}) \times (1.20 \times 10^{-6})}{0.250} \). Calculating this yields \( U = 0.198 \ \text{J} \).

Step 2: Apply Conservation of Energy for Released Charge (i)

When the charge is released, its kinetic energy comes from the change in electric potential energy. Initially, all energy is electric potential energy \( U_i = 0.198 \ \text{J} \). At \( r = 0.500 \text{ m} \), the new potential energy \( U_f = \frac{kQq}{0.500} \). Calculate \( U_f \) and use \( \Delta U = U_i - U_f = \frac{1}{2} mv^2 \) to find the speed \( v \). Plug values into \( v = \sqrt{\frac{2(U_i - U_f)}{m}} \).

Step 3: Calculate Final Electric Potential Energy (i)

For \( r = 0.500 \text{ m} \), substitute into the potential energy formula: \( U_f = \frac{(8.99 \times 10^9) \times (4.60 \times 10^{-6}) \times (1.20 \times 10^{-6})}{0.500} = 0.099 \ \text{J} \).

Step 4: Calculate Speed at 0.500 m

From \( \Delta U = 0.198 \ \text{J} - 0.099 \ \text{J} = 0.099 \ \text{J} \), use \( \Delta U = \frac{1}{2} mv^2 \) to compute speed. Solve for \( v \): \( v = \sqrt{\frac{2 \times 0.099}{2.80 \times 10^{-4}}} \approx 26.6 \ \text{m/s} \).

Step 5: Repeat for Other Distances

For \( r = 5.00 \ \text{m} \), calculate the new \( U_f \) and find speed with the same method. Then, for \( r = 50.0 \ \text{m} \), calculate \( U_f = 0 \ \text{J} \) as it nears zero, and calculate the speed again. The final speeds calculated would be \( v \approx 57.8 \ \text{m/s} \) for \( r = 5.00 \text{ m} \) and \( v \approx 59.7 \ \text{m/s} \) for \( r = 50.0 \text{ m} \).

Key Concepts

Point Charges

In the world of electromagnetism, point charges are considered to be charged particles with negligible size. These small, but mighty, charges are often assumed to be at a single point in space, making it easier to study their interactions.
Point charges can either be positive, such as protons, or negative, like electrons.
When two point charges are in proximity, they exert forces on each other. This force can be attractive or repulsive, depending on the types of charges (opposite charges attract, while like charges repel). The simplicity of considering charges as 'point-like' allows us to use mathematical expressions to predict their behavior.

• Helps in simplifying complex systems
• Used in various calculations (e.g., electric fields, potentials)

In the given exercise, we have two point charges with known values. These charges generate electric potential energy due to their positions relative to each other, a concept that plays a crucial role in many physics problems.

Coulomb's Law

Coulomb's Law is a fundamental principle of electrostatics that describes the force between two point charges.
This law states that the electric force (\( F \)) between two stationary point charges is directly proportional to the product of the magnitudes of the charges and inversely proportional to the square of the distance between them.
The formula for Coulomb's Law is:\[F = \frac{k \cdot |Q \cdot q|}{r^2}\]where:

• \( k \) is Coulomb's constant (\(8.99 \times 10^9 \text{ Nm}^2/\text{C}^2 \))
• \( Q \) and \( q \) are the magnitudes of the charges
• \( r \) is the distance between the centers of the two charges

This law allows us to calculate not only the force but also to determine the electric potential energy (\( U = \frac{kQq}{r} \)) between two charges.
By understanding Coulomb's Law, we can solve the initial part of the problem, where the potential energy is required when the charges are at a specified distance.

Conservation of Energy

The principle of conservation of energy dictates that energy cannot be created or destroyed. It can only transfer or change forms.
This conservation law is incredibly useful in physics.
In our exercise, the conservation of energy tells us how the electric potential energy (\( U \)) is converted into kinetic energy (\( KE \)) as the point charge is released.

• Initial potential energy \( U_i \) transforms into kinetic energy \( KE \) when released.
• Given by the equation: \[\Delta U = U_i - U_f = \frac{1}{2} mv^2\]

This equation helps us find the speed (\( v \)) of the released charge after it moves a certain distance.
By applying the conservation of energy, we determine how fast the charge moves at different points, using the initial potential energy and the calculated potential energy at each distance (e.g., 0.500 m, 5.00 m, etc.).