Question

A tiny sphere with a charge of q = +8.0 µC is attached to a spring....

A tiny sphere with a charge of q = +8.0 µC is attached to a spring. Two other tiny charged spheres, each with a charge of ❝4.0-µC, are placed in the positions shown in the figure, in which b = 4.5 cm. The spring stretches 5.0 cm from its previous equilibrium position toward the two spheres. Calculate the spring constant.

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Answer #1

spring constant F=-kx where x is the displacement in meters.

Using pythagorean angle a=61 degrees.

E= kQ1Q2/r^2

where k= 9e9
Q1=8.2e-6
Q2=-4.0e-6
r=.042 (found by pythagorean's theorum)

This gives you 167.34 which you need to multiply 167.34sin61 which gives you 146.35. multiply that by 2 since you have two charges. So your force equals 292.71.
Now solve for th spring contant F=-kx
292.71=-k(.05m)
k=5854.35

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Answer #2

Here is what I solved before, please modify the figures as per your question. Please let me know if you have further questions. Ifthis helps then kindly rate 5-stars.

A tiny sphere with a charge of q= +8.8

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