Electrical Force Decreased by 1/4 When Particles Moved Half as Far Apart

AI Thread Summary
The discussion centers on understanding how the electrical force between charged particles changes when the distance between them is altered, specifically when the distance is halved. According to Coulomb's Law, the force is inversely proportional to the square of the distance. When the distance is halved, the new distance squared becomes (1/2)^2, which is 1/4, leading to an increase in force by a factor of 4. Thus, the electrical force increases fourfold when the particles are moved half as far apart. This highlights the critical role of distance in determining electrical force.
weemanbran
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Homework Statement


The magnitude of electrical force between a pair of charged particles is _____ as much when the particles are moved half as far apart.


Homework Equations





The Attempt at a Solution



I understand that Coulomb's Law is.

F = k q1 q2 / d^2

So is the answer just putting in the distance which is 1/2 and squaring it? So is it just 1/2^2?
 
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Be careful -- the distance between particles appears in the denominator of Coulomb's Law. So if the distance is halved, what happens to this ratio?
 
If the distance distance is halved then the ratio would increase... I just don't know by how much...
 
Will the force increase or decrease? and by what factor? Hint: What is (\frac{1}{2})^2?
 
konthelion said:
Will the force increase or decrease? and by what factor? Hint: What is (\frac{1}{2})^2?

I think you mean \frac{1}{(\frac{1}{2})^2} ; the halved distance is in the denominator for Coulomb's Law...
 
Thread 'Variable mass system : water sprayed into a moving container'

Thread 'Variable mass system : water sprayed into a moving container'

Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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