Electric field problem -- Repulsive force between two charged spheres

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The discussion centers on the repulsive force between two charged spheres that touch and then separate, leading to a change in force. It is noted that the charges equalize upon contact, resulting in an average charge of 2.5 nC. The challenge lies in determining the original charges of the spheres before they touched. Additionally, the size of the spheres is important, as the problem assumes they are much smaller than 20 cm in radius to avoid complications in charge distribution. Understanding these factors is crucial for accurately solving the electric field problem.
Dezzi
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Homework Statement
Two identical spheres have charges q1 and q2. When they are 20 cm apart, the replusive force between them is 1.35×10^‐4N. After they touched together and separated once again to 20cm, the replusive force between them is 1.406×10^-4N. FIND q1 and q2. (Ans 20nC; 30 nC)
Relevant Equations
F=kQ1Q2/r^2
I attempt to solve the problem in the picture below.
 

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Why is the force different the second time?
 
PeroK said:
Why is the force different the second time?
The question say " they touched together and separated once again"
 
Dezzi said:
The question say " they touched together and separated once again"
So, what changed?
 
PeroK said:
So, what changed?
I guess the charges get equalised. So, you've worked out that average charge: ##2.5 nC##, which looks right.

The hard bit is to figure out what the original charges were. Any ideas?
 
A quibble: since the spheres are evidently conductors, the question ought to specify that the spheres are much smaller than 20cm radius. Otherwise the charge distributions complicate matters.
 
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Thread 'Chain falling out of a horizontal tube onto a table'

Thread 'Chain falling out of a horizontal tube onto a table'

My attempt: Initial total M.E = PE of hanging part + PE of part of chain in the tube. I've considered the table as to be at zero of PE. PE of hanging part = ##\frac{1}{2} \frac{m}{l}gh^{2}##. PE of part in the tube = ##\frac{m}{l}(l - h)gh##. Final ME = ##\frac{1}{2}\frac{m}{l}gh^{2}## + ##\frac{1}{2}\frac{m}{l}hv^{2}##. Since Initial ME = Final ME. Therefore, ##\frac{1}{2}\frac{m}{l}hv^{2}## = ##\frac{m}{l}(l-h)gh##. Solving this gives: ## v = \sqrt{2g(l-h)}##. But the answer in the book...
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