Confused about this force calulation in Resnick Halliday Krane Electrostatics

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KnightTheConqueror
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TL;DR Summary
Resnick Halliday Krane Electrostatics Sample Problem 25-7 doubt
I'm confused in the calculation for R/2. The author took in account that the charge will change by a factor 1/8. But how does it show that the coloumbic force will become 1/8th. The distance will also reduce by half shouldn't that also be taken into account? Or am I missing something here?
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KnightTheConqueror said:
TL;DR Summary: Resnick Halliday Krane Electrostatics Sample Problem 25-7 doubt

I'm confused in the calculation for R/2. The author took in account that the charge will change by a factor 1/8. But how does it show that the coloumbic force will become 1/8th. The distance will also reduce by half shouldn't that also be taken into account? Or am I missing something here?View attachment 336703
Been there, done that. You are correct and the book is wrong. See discussion here.
https://www.physicsforums.com/threa...rted-on-an-electron-inside-a-nucleus.1052346/
 
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I agree with you - they've correctly calculated the charge ratio and forgotten the distance ratio.

Edit: beaten to it, I see.
 
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:welcome:
 
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KnightTheConqueror said:
Thank you, i just started electromagnetism and i got disappointed that perhaps I'm not getting proper grasp of the concepts. Thanks for clarifying the book is wrong
I wish you success with your studies. If you need more help, you know where to find it.
 

1. How do I calculate the force between two charged particles in electrostatics?

To calculate the force between two charged particles in electrostatics, you can use Coulomb's Law. The formula is F = k * (q1 * q2) / r^2, where F is the force, k is the electrostatic constant, q1 and q2 are the charges of the particles, and r is the distance between the particles.

2. What is the electrostatic constant in the force calculation?

The electrostatic constant, denoted by k, is a physical constant that appears in Coulomb's Law. It has a value of approximately 8.9875 x 10^9 N m^2/C^2 in a vacuum. This constant is used to relate the force between two charged particles to their charges and the distance between them.

3. How does the distance between charged particles affect the force calculation?

The force between charged particles is inversely proportional to the square of the distance between them. This means that as the distance between the particles increases, the force decreases. Conversely, as the distance decreases, the force increases. This relationship is important to consider when calculating the force in electrostatics.

4. Can the force between charged particles be attractive or repulsive?

Yes, the force between charged particles can be either attractive or repulsive, depending on the signs of the charges. Like charges (positive-positive or negative-negative) repel each other, while opposite charges (positive-negative) attract each other. The magnitude of the force is the same in both cases, but the direction is determined by the signs of the charges.

5. What are the units of force in electrostatics calculations?

In electrostatics calculations, the unit of force is the newton (N). This unit is derived from the fundamental units of mass, length, and time in the International System of Units (SI). When calculating the force between charged particles using Coulomb's Law, the result will be in newtons, representing the strength of the electrostatic interaction between the particles.

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