I'm having a bit of a conundrum:
Two point charges are separated by 5 cm. The attractive force
between them is 20 N. Find the force between them when
they are separated by 10 cm.

Question

I'm having a bit of a conundrum: 
Two point charges are separated by 5 cm. The attractive force
between them is 20 N. Find the force between them when
they are separated by 10 cm.

anonymous · Answer

Coulomb's law states that $$F= \frac{ C_1*C_2 }{ (D)^2 }$$

anonymous · Answer

The solution turns to be that the Coulomb force decreases inversely to the square of the
distance. Since the distance increased twice, the force will decrease four times, so 20N/4=5N.

anonymous · Answer

I need some explaining of that matter that the "the distance increased twice, the force will decrease four times,"

anonymous · Answer

Whoop, figured out my own question!
I took the 10 and squared it getting 100. I suppose i could cancel out the cm and get 5N.

calculusfunctions · Answer

The force between two electrical charges is directly proportional to the product of their charges and inversely proportional to the square of the distance between them. We could invoke a simple algebraic approach to the problem.$$F _{e}=\frac{ kq _{1}q _{2} }{ r ^{2} }$$where  k  is the constant of proportionality. Now, when they're 5 cm apart, the force between them is 20 N. Hence$$20=\frac{ kq _{1}q _{2} }{ 5^{2} }$$ Hence multiplying this equation by 25 yields$$500=kq _{1}q _{2}$$Thus if r = 10 cm, then$$F _{e}=\frac{ kq _{1}q _{2} }{ 10^{2} }$$However$$kq _{1}q _{2}=500$$from equation 1. Thus$$F _{e}=\frac{ 500 }{ 100 }=5$$∴ when they are 10 cm apart, they have a force of  5 N  between them.

anonymous · Answer

Well.... that makes absolutely more sense than my primitive notion. My brain thanks you kindly!

calculusfunctions · Answer

You're very welcome! I'm a mathematician and a mathematics teacher. Therefore I invoke math into everything! LOL