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Chapter 1: Problem 44

A small sphere with mass \(9.00 \mu \mathrm{g}\) and charge \(-4.30 \mu \mathrm{C}\) is moving in a circular orbit around a stationary sphere that has charge \(+7.50 \mu \mathrm{C}\). If the speed of the small sphere is \(5.90 \times 10^{3} \mathrm{~m} / \mathrm{s}\), what is the radius of its orbit? Treat the spheres as point charges and ignore gravity.

Short Answer

Expert verified
The orbit radius is approximately 0.209 meters.

Step by step solution

01

Understand the problem

We need to find the radius of the circular orbit of a small charged sphere moving around a stationary charged sphere. The forces acting on the moving sphere come from the electrostatic attraction between the charges.
02

Establish the relevant formula

The centripetal force required to keep the small sphere in a circular orbit is provided by the electrostatic force. Therefore, the magnitude of the electrostatic force (given by Coulomb's Law) must equal the centripetal force. We use the formula: \[ F = \frac{k \cdot |q_1 \cdot q_2|}{r^2} = \frac{m \cdot v^2}{r} \] where \( k \) is Coulomb's constant \(8.99 \times 10^9 \frac{Nm^2}{C^2}\), \( q_1 \) and \( q_2 \) are the charges of the spheres, \( m \) is the mass of the moving sphere, \( v \) is its velocity, and \( r \) is the radius of the orbit.
03

Substitute given values into the formula

We need to rearrange the centripetal and electrostatic force formula to solve for \( r \): \[ \frac{k \cdot |q_1 \cdot q_2|}{r} = m \cdot v^2 \] thus \[ r = \frac{k \cdot |q_1 \cdot q_2|}{m \cdot v^2} \]Substitute the given values: \( k = 8.99 \times 10^9 \frac{Nm^2}{C^2} \), \( q_1 = 4.30 \times 10^{-6} C \), \( q_2 = 7.50 \times 10^{-6} C \), \( m = 9.00 \times 10^{-9} kg \), \( v = 5.90 \times 10^3 m/s \).
04

Perform the calculation

Plug the values into the rearranged formula: \[ r = \frac{8.99 \times 10^9 \times |4.30 \times 10^{-6} \times 7.50 \times 10^{-6}|}{9.00 \times 10^{-9} \times (5.90 \times 10^3)^2} \]This simplifies to:\[ r = \frac{8.99 \times 10^9 \times 3.225 \times 10^{-11}}{9.00 \times 10^{-9} \times 3.481 \times 10^7} \]Calculating the above gives:\[ r \approx 0.209 \text{ meters} \]
05

Interpret the result

The calculated radius of the orbit is approximately 0.209 meters. This means that the path followed by the small sphere around the stationary sphere due to the electrostatic force acting as the centripetal force has a radius of about 0.209 meters.

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Key Concepts

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

Coulomb's Law
Coulomb's Law is pivotal in understanding electrostatic forces between charged objects. It quantifies the amount of force between two stationary, electrically charged particles. The law states that the electrostatic force (\( F \)) between two charges is directly proportional to the product of the magnitudes of the charges (\( |q_1 \, ext{and} \, q_2| \)) and inversely proportional to the square of the distance (\( r \)) between them. The formula for Coulomb's Law is given by:\[F = \frac{k \cdot |q_1 \cdot q_2|}{r^2}\]Where \( k \) is Coulomb's constant (\( 8.99 \times 10^9 \text{ N}\cdot\text{m}^2/\text{C}^2 \)). This force is attractive if the charges are of opposite sign and repulsive if they are of the same sign.
  • Directly proportional: The force increases with greater charge magnitudes.
  • Inversely proportional: The force decreases as the distance between charges increases.
Understanding this law helps in analyzing situations where electric charges interact, such as when two charged spheres attract or repel each other.
Centripetal Force
When an object moves in a circular path, it continuously changes its direction. To keep changing direction, the object requires a force directed towards the center of the circle. This force is known as the centripetal force. In this exercise, centripetal force plays a crucial role in maintaining the orbit of the small sphere. The formula for centripetal force (\(F_c\)) is:\[F_c = \frac{m \cdot v^2}{r}\]Where
  • \( m \) is the mass of the moving object.
  • \( v \) is its velocity.
  • \( r \) is the radius of the circular path.
The centripetal force is not a separate force but a resultant of other forces. Here, the electrostatic force provided by the stationary charged sphere acts as the centripetal force. This pulls the moving sphere towards the center of its orbit, allowing it to move in a circle. Understanding centripetal force is foundational for analyzing circular motion in physics.
Orbits Calculation
In circular motion, calculating the orbit's radius is crucial to understanding the path of a moving object. For this problem, the small sphere orbits around a stationary sphere. To find the orbit's radius, we equate the centripetal force to the electrostatic force produced by Coulomb's Law, as these forces balance each other to maintain circular motion:\[\frac{k \cdot |q_1 \cdot q_2|}{r} = m \cdot v^2\]From this, rearrange to solve for the radius \( r \):\[r = \frac{k \cdot |q_1 \cdot q_2|}{m \cdot v^2}\]Substituting known values allows us to compute the radius. By accurately inserting the charge values, constants, mass, and velocity into the formula, we find that the radius is approximately 0.209 meters.
  • This calculation relies on understanding and applying both Coulomb's Law and centripetal force.
  • It's fundamental for problems involving charges and motion, like planetary systems or atomic structures.
This approach helps in predicting behaviors in systems where circular orbits due to electrostatic forces are present.

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Most popular questions from this chapter

Positive charge \(Q\) is distributed uniformly along the positive \(y\)-axis between \(y=0\) and \(y=a\). A negative point charge \(-q\) lies on the positive \(x\)-axis, a distance \(x\) from the origin (Fig. P1.48). (a) Calculate the \(x\) - and \(y\)-components of the electric field produced by the charge distribution \(Q\) at points on the positive \(x\)-axis. (b) Calculate the \(x\) - and \(y\)-components of the force that the charge distribution \(Q\) exerts on \(q\). (c) Show that if \(x \gg a, F_{x} \cong-Q q / 4 \pi \epsilon_{0} x^{2}\) and \(F_{y} \cong+Q q a / 8 \pi \epsilon_{0} x^{3} .\) Explain why this result is obtained.

A proton is projected into a uniform electric field that points vertically upward and has magnitude \(E\). The initial velocity of the proton has a magnitude \(v_{0}\) and is directed at an angle \(\alpha\) below the horizontal. (a) Find the maximum distance \(h_{\max }\) that the proton descends vertically below its initial elevation. You can ignore gravitational forces. (b) After what horizontal distance \(d\) does the proton return to its original elevation? (c) Sketch the trajectory of the proton. (d) Find the numerical values of \(h_{\max }\) and \(d\) if \(E=500 \mathrm{~N} / \mathrm{C}, v_{0}=4.00 \times 10^{5} \mathrm{~m} / \mathrm{s}\), and \(\alpha=30.0^{\circ}\)

An infinite sheet with positive charge per unit area \(\sigma\) lies in the \(x y\)-plane. A second infinite sheet with negative charge per unit area \(-\sigma\) lies in the \(y z\)-plane. Find the net electric field at all points that do not lie in either of these planes. Express your answer in terms of the unit vectors \(\hat{\boldsymbol{\imath}}, \hat{\boldsymbol{J}}\), and \(\hat{\boldsymbol{k}}\).

A tiny electric dipole \(\left(\mathrm{H}_{2} \mathrm{O}\right.\) modecule) of dipole moment \(\vec{p}\) is placed at a distance \(r\) form an infinitely long wire, with its \(\vec{p}\) normal to the wire in the same plane. If the linear charge density of the wire is \(\lambda\), find the electrostatic force acting on the dipole is equal to

Two particles having charges \(q_{1}=0.500 \mathrm{nC}\) and \(q_{2}=8.00 \mathrm{n} \mathrm{C}\) are separated by a distance of \(1.20 \mathrm{~m}\). At what point along the line connecting the two charges is the total electric field due to the two charges equal to zero?
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