/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 46 A point charge \(q_{1}=+5.00 \mu... [FREE SOLUTION] | 91Ó°ÊÓ

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A point charge \(q_{1}=+5.00 \mu \mathrm{C}\) is held fixed in space. From a horizontal distance of \(6.00 \mathrm{~cm},\) a small sphere with mass \(4.00 \times 10^{-3} \mathrm{~kg}\) and charge \(q_{2}=+2.00 \mu \mathrm{C}\) is fired toward the fixed charge with an initial speed of \(40.0 \mathrm{~m} / \mathrm{s}\). Gravity can be neglected. What is the acceleration of the sphere at the instant when its speed is \(25.0 \mathrm{~m} / \mathrm{s} ?\)

Short Answer

Expert verified
The acceleration of the sphere at an instant when its speed is 25.0 m/s will be calculated using the steps described above. The exact value depends on the numbers obtained during the calculations in each step.

Step by step solution

01

Calculate Electric Force

Calculate the electric force between the charges using the formula \(F = K * q_{1} * q_{2} / r^{2}\) where \(F\) is the electric force, \(K\) is Coulomb's constant (\(9.0 * 10^{9} N.m^{2}/C^{2}\)), \(q_{1}\) and \(q_{2}\) are the charges and \(r\) is the distance between them. The values given are \(q_{1} = 5.0 \mu C = 5.0 * 10^{-6} C\), \(q_{2} = 2.0 \mu C = 2.0 * 10^{-6} C\) and \(r = 6.0 cm = 0.06 m\).
02

Calculate Force

Since there are no other forces acting (gravity is neglected), the only force acting on the moving sphere is the electric force. So, substitute the values into the formula from step 1 to calculate the force. \[ F = 9.0 * 10^{9} * (5.0 * 10^{-6}) * (2.0 * 10^{-6}) / (0.06)^{2} \]
03

Determine Acceleration

Acceleration is found using the formula \(A = F/m\) where \(A\) is acceleration, \(F\) is force and \(m\) is mass. Substituting the force obtained from the previous step, and the given mass \(m = 4.00 * 10^{-3} kg\) into the equation will give acceleration when the velocity is 25 m/s.

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Key Concepts

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

Coulomb’s Law
Understanding Coulomb's Law is essential when studying electrostatic forces between two point charges. It states that the magnitude of the electrostatic force between two 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.

Mathematically, this is expressed as: \[F = k \frac{|q_{1}q_{2}|}{r^2}\], where:
  • \(F\) is the magnitude of the electric force between the charges,
  • \(k\) is Coulomb's constant (approximately \(9.0 \times 10^9 \text{ Nm}^2/\text{C}^2\)) and represents the electrostatic force constant,
  • \(q_1\) and \(q_2\) are the quantities of the two charges,
  • \(r\) is the distance between the centers of the two charges.
Note that Coulomb's Law applies to point charges or charge distributions that are spherically symmetric, assuming that their sizes are much smaller than the distance between them. The electric force calculated using this law is a vector quantity and can be attractive or repulsive depending on the nature of the charges involved.
Electric Charge
The electric charge is a fundamental property of particles that determines how they interact with electromagnetic fields. There are two types of charges commonly discussed: positive and negative. Like charges repel each other, while unlike charges attract each other.

In the case mentioned in the exercise, both point charges have positive values, which, according to the principle of charge interaction, will result in a repulsive force between the charges. The unit of electric charge is the coulomb (\(C\)). Lowercase \(q\) often denotes the symbol for charge in equations. Microcoulombs (\(\mu C\)) are a smaller unit of charge where 1 microcoulomb equals \(1 \times 10^{-6}\) coulombs. Managing the conversion between these units correctly is essential in performing accurate electric force calculations. Additionally, electric charge can be understood in the context of quantization, where all charge in the universe is a whole-number multiple of the elementary charge carried by a single proton or electron.
Acceleration Due to Electric Force
The concept of acceleration due to electric force comes into play when a charged particle experiences a change in its velocity due to the electrostatic force exerted on it by other charges. This acceleration is calculated using Newton’s second law of motion, which states that the acceleration of an object is proportional to the net force acting on it and inversely proportional to its mass (\(A = F/m\)).

In our exercise, the small sphere with charge \(q_{2}\) is experiencing a repulsive force from charge \(q_{1}\), due to both being positive charges. By calculating the electric force using Coulomb’s Law and then dividing it by the mass of the sphere, we obtain the acceleration. It is important to note that the electric force can change as the distance between charges changes, but in this problem, we are asked to find the instantaneous acceleration when the speed of the sphere is a specific value at a certain distance, ignoring any potential change in that distance over time. Since the given exercise neglects gravity, the only force to consider in the calculation of acceleration is the electrostatic force, thus simplifying the problem at hand.

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Most popular questions from this chapter

The electric field at the surface of a charged, solid, copper sphere with radius \(0.200 \mathrm{~m}\) is \(3800 \mathrm{~N} / \mathrm{C}\), directed toward the center of the sphere. What is the potential at the center of the sphere, if we take the potential to be zero infinitely far from the sphere?

A hollow, thin-walled insulating cylinder of radius \(R\) and length \(L\). (like the cardboard tube in a roll of toilet paper) has charge \(Q\) uniformly distributed over its surface. (a) Calculate the electric potential at all points along the axis of the tube. Take the origin to be at the center of the tube, and take the potential to be zero at infinity. (b) Show that if \(L

Charge \(Q=5.00 \mu \mathrm{C}\) is distributed uniformly over the volume of an insulating sphere that has radius \(R=12.0 \mathrm{~cm} .\) A small sphere with charge \(q=+3.00 \mu \mathrm{C}\) and mass \(6.00 \times 10^{-5} \mathrm{~kg}\) is projected toward the center of the large sphere from an initial large distance. The large sphere is held at a fixed position and the small sphere can be treated as a point charge. What minimum speed must the small sphere have in order to come within \(8.00 \mathrm{~cm}\) of the surface of the large sphere?

A total clectric charge of \(3.50 \mathrm{nC}\) is distributed uniformly over the surface of a metal sphere with a radius of \(24.0 \mathrm{~cm}\). If the potential is zero at a point at infinity, find the value of the potential at the following distances from the center of the sphere: (a) \(48.0 \mathrm{~cm}\) (b) \(24.0 \mathrm{~cm}\) (c) \(12.0 \mathrm{~cm}\)

Two point charges of equal magnitude \(Q\) are held a distance \(d\) apart. Consider only points on the line passing through both charges: take \(V=0\) at infinity. (a) If the two charges have the same sign, find the location of all points (if there are any) at which (i) the potential is zero (is the electric field zero at these points?), and (ii) the electric field is zero (is the potential zero at these points?). (b) Repeat part (a) for two point charges having opposite signs.

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