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# plzz help me with this i need to know 100% correct answer?

I have two equal positive charges placed 38 cm apart. They are moved so the force between them decreases by a factor of 6 times the original force. What is the distance between them?

### 5 Answers

- Ian ILv 41 decade ago
Electrostatic repulsive forces obey the inverse square law. This means that if they experience a force F at a distance R, then the magnitude of the force is given by F = N/(R^2).

So, in this case, you start off with force F = N/(38^2) = N/1444, where N is a constant.

At what distance 'r' will F be reduced to (1/6)F?

N is constant, but F is REDUCED by a factor of 6. This means that R^2 must have INCREASED by a factor of 6.

So, 1444 x 6 = 8664 = new R^2

So, new R is sqrroot 8664 = 93 cm.

- Anonymous1 decade ago
No such distance exists.

Moving the cahrges apart to infinity will decrease the force between them to zero. This is a reduction of 1 times the original force. There can be no greater reduction.

- Anonymous1 decade ago
94.28 cm

F1=kQ²/r²

F2=kQ²/(r+x)²

(r+x)²=6r²

r²+2rx+x²=6r²

if r=38, then:

1444+72x+x²=8664

x²+72x-7220=0

x=(-72+sqrt(72²+4*7220))/2 = 56.28

So the new distance is r+x = 38+56.28 = 94.28

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