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Two small plastic spheres are given positive electrical charges. When they are 15.0 \(\mathrm{cm}\) apart, the repulsive force between them has magnitude 0.220 \(\mathrm{N}\) . What is the charge on each sphere (a) if the two charges are equal and (b) if one sphere has four times the charge of the other?

Short Answer

Expert verified
If the charges are equal, each is approximately 2.86 × 10^-7 C; if one is four times the other, they are approximately 1.43 × 10^-7 C and 5.72 × 10^-7 C.

Step by step solution

01

Understanding the Coulomb's Law

The problem involves determining electrical charges using Coulomb's Law, which is mathematically represented as \( F = k \frac{q_1 q_2}{r^2} \), where \( F \) is the Coulomb force, \( k \) is the Coulomb constant \( 8.99 \times 10^9 \, \text{N m}^2/\text{C}^2 \), \( q_1 \) and \( q_2 \) are the charges, and \( r \) is the distance between them. Given: \( F = 0.220 \, \text{N} \) and \( r = 0.15 \, \text{m} \).
02

Computing Charge for Equal Charges

If the charges are equal, let each charge be \( q \). Substitute \( q_1 = q_2 = q \) in the formula to get \( F = k \frac{q^2}{r^2} \). Rearrange for \( q \): \( q = \sqrt{\frac{F r^2}{k}} \). Plug in the values: \( q = \sqrt{\frac{0.220 \times (0.15)^2}{8.99 \times 10^9}} \approx 2.86 \times 10^{-7} \, \text{C} \).
03

Computing Charge for Non-Equal Charges

For one charge being four times the other, let \( q_1 = q \) and \( q_2 = 4q \). Substitute in Coulomb's law: \( F = k \frac{q (4q)}{r^2} \) which becomes \( F = k \frac{4q^2}{r^2} \). Rearrange for \( q \): \( q = \sqrt{\frac{F r^2}{4k}} \). Substitute the values: \( q = \sqrt{\frac{0.220 \times (0.15)^2}{4 \times 8.99 \times 10^9}} \approx 1.43 \times 10^{-7} \, \text{C} \). The other charge \( 4q = 5.72 \times 10^{-7} \, \text{C} \).

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Key Concepts

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

Electric Charges
Electric charges are fundamental properties of matter that lead to electromagnetic interactions. These charges can either be positive or negative, and they dictate the nature of forces that act between charged objects.
Charged particles can exert forces over a distance, much like how magnets work, without needing to be in direct contact. Similar to how opposite magnet poles attract whereas similar poles repel, electric charges behave in the same fashion:
  • Like charges (positive-positive or negative-negative) repel each other.
  • Opposite charges (positive-negative) attract each other.
Understanding these basic principles is crucial when dealing with problems involving electric forces, such as the one in our original exercise. Keeping these fundamentals in mind will help make sense of how forces and charges interact.
Force Between Charges
The force between electric charges is described by Coulomb's Law, which quantifies the interaction. This law states that the force (F) between two point charges is directly proportional to the product of the magnitudes of the charges (q_1 and q_2) and inversely proportional to the square of the distance (r) between them:\[F = k \frac{q_1 q_2}{r^2}\]where k is Coulomb's constant (8.99 \times 10^9 \text{ N m}^2/\text{C}^2).

The widespread application of this formula allows you to predict how strong the repulsive or attractive force is between two charges. Importantly, the square of the distance in the formula shows us that changing the distance even slightly can have a large impact on the magnitude of the force. This inverse-square law is crucial in solving problems like the one discussed, where knowing the force helps us find unknown charges when other variables are given.
Equal and Unequal Charges
When solving problems involving equal charges, the math simplifies considerably because both charges have the same value. In these cases, each charge is simply denoted as q, and the problem becomes about finding q based on given values like distance and force.
In our exercise, when the charges are equal, the formula simplifies to:\[F = k \frac{q^2}{r^2}\]This means the force is calculated based on one variable square, reducing complexity.
For unequal charges, the scenario shifts slightly. If one charge is a multiple of the other, extra care is needed. In the problem, one charge is four times the other, thus taking on the forms q and 4q. The relationship becomes:\[F = k \frac{4q^2}{r^2}\]By understanding the relationships between charges, these complex problems become more manageable. The operations involve more algebra, but the principles and steps remain grounded in fundamental physics.
Problem-Solving Steps
To solve any physics problem related to Coulomb's Law effectively, a structured approach is essential. Let’s outline the steps based on the given problem:
  • Identify knowns and unknowns: Begin by gathering what is already given in the problem statement: the distance ( r ) between the charges and the force ( F ) are known here.
  • Set up Coulomb’s Law: Use the law as your equation base. Choose appropriate formulas based on whether the charges are equal or unequal.
  • Solve for the unknown: Rearrange the formula to solve for the unknown variable. With equal charges, rearrange to find q . With unequal charges, you must consider the relationship between the charges, like q and 4q .
  • Substitute and calculate: Insert all known values into the rearranged formula and carefully compute to find your answer.
Following these steps not only provides discipline but also reduces errors, ensuring you've correctly applied physical laws to reach a conclusion. Efficient problem-solving in physics builds confidence and deepens comprehension of fundamental concepts like electrodynamics.

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Most popular questions from this chapter

(a) What must the charge (sign and magnitude) of a \(1.45-\mathrm{g}\) particle be for it to remain stationary when placed in a downward- directed electric field of magnitude 650 \(\mathrm{N} / \mathrm{C} 2\) (b) What is the magnitude of an electric field in which the electric force on a proton is equal in magnitnde to its weight?

Point charges \(q_{1}=-4.5 \mathrm{nC}\) and \(q_{2}=+4.5 \mathrm{nC}\) are separated by \(3.1 \mathrm{mm},\) forming an electric dipole. (a) Find the electric dipole moment (magnitude and dircction). (b) The charges are in a uniform electric field whose direction makes an angle of \(36.9^{\circ}\) with the line connecting the charges. What is the magnitude of this field if the torque exerted on the dipole has magnitude \(7.2 \times 10^{-9} \mathrm{N} \cdot \mathrm{m} ?\)

(a) An electron is moving east in a uniform electric field of 1.50 \(\mathrm{N} / \mathrm{C}\) directed to the west. At point \(A,\) the velocity of the electron is \(4.50 \times 10^{5} \mathrm{m} / \mathrm{s}\) toward the east. What is the speed of the electron when it reaches point \(B, 0.375 \mathrm{m}\) east of point \(A ?(\mathrm{b}) \mathrm{A}\) proton is moving in the uniform electric field of part (a). At point \(A .\) the velocity of the proton is \(1.90 \times 10^{4} \mathrm{m} / \mathrm{s}\) , east. What is the speed of the proton at point \(B ?\)

A straight, nonconducting plastic wire 8.50 \(\mathrm{cm}\) long carries a charge density of \(+175 \mathrm{nC} / \mathrm{m}\) distributed uniformly along its length. It is lying on a horizontal tabletop. (a) Find the magnitude and direction of the electric field this wire produces at a point 6.00 \(\mathrm{cm}\) directly above its midpoint. (b) If the wire is now bent into a circle lying flat on the table, find the magnitude and direction of the electric field it produces at a point 6.00 \(\mathrm{cm}\) directly above its center.

Two horizontal, infinite, plane sheets of charge are separated by a distance \(d\) . The lower sheet has negative charge with uniform surface charge density \(-\sigma<0 .\) The upper sheet has positive charge with uniform surface charge density \(\sigma>0 .\) What is the electric field (magnitude, and direction if the field is nonzero) (a) above the upper sheet, (b) below the lower sheet, (c) between the sheets?

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