How Do You Calculate the Net Force on Particles in a Line with Varied Charges?

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To calculate the net force on the middle and right particles in a line with varied charges, one must apply Coulomb's Law, which relates the force between two charged particles to their charges and the distance between them. For the middle particle, the forces from the left and right particles are calculated, resulting in a net force of -0.0279 N acting towards the left. For the right particle, the forces from both the left and middle particles yield a net force of -0.0052 N, also directed towards the left. It is essential to systematically identify and sum the individual forces acting on each particle to determine the net force accurately. Understanding the direction and magnitude of these forces is crucial for solving problems involving charged particles.
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i have a 2 specific problems, I am having trouble, if anyone can help me figure them out,

1) Three particles are placed in a line The left particle has a charge of
(-67 x 10^-6 C)

The middle partilce has a charge of (+45 x 10^-6C) and is 72cm from each of the others, and the right has a cahrge of (-83 x 10^-6)
{FIND THE NET FORCE ON MIDDLE AND RIGHT PARTICLE??) :eek:
 
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First of all, BE SYSTEMATIC!
You must solve 2 problems:
1) Calculate the net force on the middle particle
2) Calculate net force on the right particle

i)
Before you attempt to solve either of these problems, you must clarify yourself about what is meant with "net force".

If an object is subjected to several forces at the same time (for example by interacting with other objects), the "net force" on the object is the sum of all the (individual) forces acting upon it.

ii) Now, you must consider:
What (individual) forces acts upon the middle particle?
(This will be the first step in solving problem 1))
I assume you know about Coulomb's law, which states what force a single charged particle experiences from another charged particle.
Clearly, the other particle experiences a force of equal magnitude, but opposite direction.

Now, Coulomb's law is valid for every pair of particles, and the expression for the (individual) force(s) associated with these 2 particles doesn't change a bit just because a third particle is present!

That is, if you have a system of particles, and you wish to calculate the net force on, say the first particle, you must sum together all the (individual) forces that particle experiences as being seen as a member of a pair of particles!
If you have 3 particles, and wish to calculate the force on particle 1, add together the force acting upon it from particle 2 (according to Coulomb's law stated for particles 1 and 2),
AND the force acting upon it from particle 3 (according to Coulomb's law stated for particles 1 and 3)

Write down which force acts upon which particle from which other particle like:
Force on particle 1 from particle 2: , and so on.
(Hint: For your problem, there is a total of 6 such forces, you will need only 4 of them to solve your problems)

Be particularly careful to use correct expressions for charges, distances and the direction of the forces!
 
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To determine the net force on the middle and right particles, we need to use Coulomb's Law, which states that the force between two charged particles is directly proportional to the product of their charges and inversely proportional to the square of the distance between them.

First, let's label the particles as A, B, and C, with A being the left particle, B being the middle particle, and C being the right particle.

To find the net force on B, we need to find the individual forces acting on it from A and C. The force from A on B can be calculated using the formula F = k(q1q2)/r^2, where k is the Coulomb's constant (9x10^9 Nm^2/C^2), q1 is the charge of A and q2 is the charge of B, and r is the distance between them. Plugging in the values, we get:

FAB = (9x10^9)(-67x10^-6)(45x10^-6)/(0.72)^2 = -0.0125 N

Similarly, the force from C on B can be calculated as:

FCB = (9x10^9)(-83x10^-6)(45x10^-6)/(0.72)^2 = -0.0154 N

Since the forces are acting in opposite directions, we need to add them together to get the net force on B:

FBnet = FAB + FCB = (-0.0125) + (-0.0154) = -0.0279 N

Therefore, the net force on the middle particle is -0.0279 N.

To find the net force on the right particle, we can use the same process. The force from A on C can be calculated as:

FAC = (9x10^9)(-67x10^-6)(-83x10^-6)/(1.44)^2 = 0.0102 N

And the force from B on C can be calculated as:

FBC = (9x10^9)(45x10^-6)(-83x10^-6)/(0.72)^2 = -0.0154 N

Adding these two forces together, we get the net force on C:

FCnet = FAC + FBC = (0.0102) + (-0.0154) = -0.0052 N
 
The book claims the answer is that all the magnitudes are the same because "the gravitational force on the penguin is the same". I'm having trouble understanding this. I thought the buoyant force was equal to the weight of the fluid displaced. Weight depends on mass which depends on density. Therefore, due to the differing densities the buoyant force will be different in each case? Is this incorrect?

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