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Chapter 18: Problem 23

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 \(\mathrm{m} ?\)

Short Answer

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

Step by step solution

01

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 \( U = 0.198 \ \text{J} \).
02

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 \( v = \sqrt{\frac{2(U_i - U_f)}{m}} \).
03

Calculate Final Electric Potential Energy (i)

For \( r = 0.500 \text{ m} \), substitute into the potential energy formula: \( U_f = \frac{(8.99 \times 10^9) \times (4.60 \times 10^{-6}) \times (1.20 \times 10^{-6})}{0.500} = 0.099 \ \text{J} \).
04

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 \( v \): \( v = \sqrt{\frac{2 \times 0.099}{2.80 \times 10^{-4}}} \approx 26.6 \ \text{m/s} \).
05

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 \( r = 50.0 \text{ m} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

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.).

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Most popular questions from this chapter

(a) How much charge does a battery have to supply to a 5.0\(\mu \mathrm{F}\) capacitor to create a potential difference of 1.5 \(\mathrm{V}\) across its plates? How much energy is stored in the capacitor in this case? (b) How much charge would the battery have to supply to store 1.0 \(\mathrm{J}\) of energy in the capacitor? What would be the potential across the capacitor in that case?

Electrical sensitivity of sharks. Certain sharks can detect an electric field as weak as 1.0\(\mu \mathrm{V} / \mathrm{m} .\) To grasp how weak this field is, if you wanted to produce it between two parallel metal plates by connecting an ordinary 1.5 \(\mathrm{V}\) A battery across theseplates, how far apart would the plates have to be?

A \(\mathrm{A} 20.0 \mu \mathrm{F}\) capacitor is charged to a potential difference of 800 \(\mathrm{V} .\) The terminals of the charged capacitor are then connected to those of an uncharged 10.0\(\mu \mathrm{F}\) capacitor. Compute (a) the onginal charge of the system, (b) the final potential difference across each capacitor, (c) the final energy of the system, and (d) the decrease in energy when the capacitors are connected.

A potential difference of 4.75 \(\mathrm{kV}\) is established between parallel plates in air. If the air becomes ionized (and hence electrically conducting) when the electric field exceeds \(3.00 \times 10^{6} \mathrm{V} / \mathrm{m},\) what is the minimum separation the plates can have without ionizing the air?

Electronic flash units for cameras contain a capacitor for storing the energy used to produce the flash. In one such unit, the flash lasts for \(\frac{1}{675}\) s with an average light power output of \(2.70 \times 10^{5} \mathrm{W}\) (a) If the conversion of electrical energy to light is 95\(\%\) efficient (the rest of the energy goes to thermal energy), how much energy must be stored in the capacitor for one flash? (b) The capacitor has a potential difference between its plates of 125 \(\mathrm{V}\) when the stored energy equals the value calculated in part (a). What is the capacitance?
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