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Consider this change to that situation. You charge the balls so that they hang a distance r apart. Then you step out to get a drink of water, and when you return, you find the distance between the pith balls is half what it was before you got a drink. In terms of the length L, the charge Q, and the original angle θ, find the new charge on the pith balls and the new angle at which they hang. To receive credit, you must show your work. (10 pts each)

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Answer:

Θ =tan⁻¹ (4KQ²/mgr²), Q = r[mgtanΘπ∈₀][tex]\frac{1}{2\\}[/tex]

Step-by-step explanation:

initially the angle Θ=0° ,the vertical forces were equal to product of mass and gravity(m*g) and there was no horizontal or lateral force in action. But after the displacement of balls new forces are induced.

X-Axis:

Fe = TsinΘ

[KQ²/(r/2)²] = TsinΘ         where r₁=r/2, r₁ = new distance

(4KQ²/r²) = TsinΘ

Y-Axis

TcosΘ = mg

As we know that tanΘ=sinΘ/cosΘ

We have, tanΘ = 4KQ²/mgr²

By adjusting this equation and putting K=1/4π∈₀ we get,

Q = r[mgtanΘπ∈₀][tex]\frac{1}{2\\}[/tex]

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