In Example 21.4, suppose the point charge on the y -axis at y=-0.30 m has negative charge -2.0 μ

In Example 21.4, suppose the point charge on the y -axis at...

Key Concepts

Coulomb's Law

Coulomb's Law states that the force between two point charges is directly proportional to the product of the charges and inversely proportional to the square of the distance between them. This principle is fundamental in electrostatics and is used to calculate how the magnitude of the electric force changes with distance and the amount of charge present, regardless of the charge sign.

Superposition Principle

The Superposition Principle in electrostatics states that the net force on any charge is the vector sum of the forces exerted by all other individual charges. This means that each pair-wise interaction is calculated independently using Coulomb's Law, and then these forces are added together to find the overall force on the charge in question.

Vector Addition of Forces

Since forces are vector quantities, they have both magnitude and direction. When calculating the net force on a charge, each contributing force must be resolved into its components (typically along mutually perpendicular axes), and then these components are summed. This allows for a complete description of the force's magnitude and the direction in which it acts.

Effect of Charge Polarity

The polarity or sign of a charge determines whether the interaction between two charges is attractive or repulsive. When one of the charges changes sign, the direction of the force between it and another charge may reverse, altering the net force on a charge. Understanding how the change in charge polarity affects the overall force is critical when comparing different charge configurations in an electrostatic problem.

Transcript

00:01 Question number 16 wants us to find the net force on the charge q from these other two charges on the y axis.

00:10 The magnitudes of each of the three charges are as follows.

00:15 Q1 has a charge of two microculems.

00:20 Q2 has a charge of minus two microculems.

00:25 And the charge q has a charge of a charge of of four microculems.

00:33 The way these charges are oriented with respect to each other, in addition to the magnitudes of their charges, will affect the forces they exert on one another.

00:44 Charge q1 is located on the y axis at a distance of 0 .3 meters from the origin.

00:51 Charge q2, meanwhile, is located 0 .3 meters below the x -axis, and the charge q is located on the x -axis at 0 .4 meters from the origin.

01:07 Now, the distance between q1 and q, that will be most useful in helping us to determine the force exerted upon the charge q, will be this distance here, 0 .5.

01:21 Meters.

01:23 The same is true for the charge q2.

01:26 The force it exerts on the charge q will be over this distance, which is also 0 .5 meters.

01:37 Now f sub 1, the force exerted by q1 on q will be in this direction at some angle alpha below the x -axis.

01:49 To determine the magnitude of f1, we need to find the value of k times q1 times q over r squared.

02:06 We have values for all of these terms, so let's go ahead and plug them in.

02:11 K is the proportionality constant.

02:14 It is 9 times 10 to the 9 newtons times meters squared per coulum q1's charge is two microculems or 2 times 10 to the minus 6 coolums and q's charge is 4 times 10 to the minus 6 coolums and this is acting over r squared which is point 5 meters squared.

02:48 When we work out our calculations, we find that f1 has a value of 0 .29 newtons.

02:56 Now let's find the magnitude of f2, the force exerted on q by the charge q2.

03:04 That will be equal to k times q2 times q2 divided by r squared.

03:13 That will be equal to 9 times 10 to the 9 newtons times meters squared per coulum, times minus 2 times 10 to the minus 6 coolums, times 4 times 10 to the minus 6 coulums divided by 0 .5 meters squared...

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