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TEA [102]
3 years ago
13

Force F acts between two charges, q1 and q2, separated by a distance d. If q1 is increased to twice its original value and the d

istance between the charges is also doubled, what is the new force acting between the charges in terms of F? F F F 2F
Physics
2 answers:
Step2247 [10]3 years ago
9 0
Okay, haven't done physics in years, let's see if I remember this.

So Coulomb's Law states that F = k \frac{Q_1Q_2}{d^2} so if we double the charge on Q_1 and double the distance to (2d) we plug these into the equation to find

<span>F_{new} = k \frac{2Q_1Q_2}{(2d)^2}=k \frac{2Q_1Q_2}{4d^2} = \frac{2}{4} \cdot k \frac{Q_1Q_2}{d^2} = \frac{1}{2} \cdot F_{old}</span>

So we see the new force is exactly 1/2 of the old force so your answer should be \frac{1}{2}F if I can remember my physics correctly.

Lesechka [4]3 years ago
5 0

The Correct answer is 1/2F...

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Area \,\,Trapezoid=(\left[B+b]\,(H/2)\\displacement= \left[(40-0)+(30-10)\right] \,(20/2)=600\,\,m

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3 years ago
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Answer:

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-Q + y = 0

∴ y = +Q

The charge on the inner shell, y = +Q

5 0
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