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E=kq/r^2

So, when the distance id doubled, the change is taken down by a factor of 1/4.

In other words, if the distance was 3, and then 6, this is what would happen:

E=kq/(3)^2=kq/9

and then when r = 6,

E=kq(6)^2=kq/36=kq/9*1/4

Thus, reduced by a factor of four or multiplied by 1/4

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14y ago
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13y ago

the electrostatic force between the two charges is inversely proportional to the square of the distance between them, which in this case means the force goes up four times.

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12y ago

The force is increased by a factor of 4, 4f. F = k/r2

halving then F=k/(.5r)2 = 4k/r2.

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12y ago

The force is multiplied by 4.

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Anonymous

Lvl 1
4y ago

Thg

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Q: Determine the amount by which the electric force between two charges is increased when the distance between the charge is halved?
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