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Two small spheres with mass m = 15.0 g are hung by silk threads of length L = 1.20 m from a common point (Fig. P21.62). When the spheres are given equal quantities of negative charge, so that q1 = q2 = q, each thread hangs at u = 25.0° from the vertical. (a) Draw a diagram showing the forces on each sphere. Treat the spheres as point charges. (b) Find the magnitude of q. (c) Both threads are now shortened to length L = 0.600 m, while the charges q1 and q2 remain unchanged. What new angle will each thread make with the vertical? (Hint: This part of the problem can be solved numerically by using trial values for 𝛉 and adjusting the values of 𝛉 until a self-consistent answer is obtained.)

Short Answer

Expert verified

(a)


(b) The magnitude of q is 2.8×10-6C

(c) If threads are now shortened to length L = 0.600 m, while the charges q1 and q2 remain unchanged, the new angle each thread will make with the vertical is 39.5°.

Step by step solution

01

Forces on each ball

The diagram shows forces on each ball

02

Equilibrium condition

The system is in equilibrium so the net force in each direction is zero

Fx=0

Also, in y-direction the net force is zero

Tcosθ=mg

From the above equations get

F=mgtanθ

The repulsive force F is given by

F=kq22d2=kq22Lsinθ2

03

Substitution

From step 2, we get

q=2Lsinθ2mgtanθK=2×1.2sin2520.0159.8tan259×109=2.8×10-6C

Therefore, the magnitude of q is 2.8×10-6C

04

Direction

From step 2, we get

sinθ2tanθ=q2k4L2mg=0.33

θ=39.5°will satisfy the relation by trial method (adjusting the value ofθ)

sin39.52tan39.50.333Therefore, if threads are now shortened to length L = 0.600 m, while the charges q1 and q2 remain unchanged, the new angle each thread will make with the vertical is 39.5°.

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