Chapter 4: Q63P (page 717)

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 $2.8 \times 10^{- 6} C$

(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

Forces on each ball

The diagram shows forces on each ball

Equilibrium condition

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

$\sum_{} F_{x} = 0$

Also, in y-direction the net force is zero

$T \cos θ = m g$

From the above equations get

$F = m g \tan θ$

The repulsive force F is given by

$F = \frac{k q^{2}}{{(2 d)}^{2}} = \frac{k q^{2}}{{(2 L \sin θ)}^{2}}$

Substitution

From step 2, we get

$q = \sqrt{\frac{{(2 L \sin θ)}^{2} m g \tan θ}{K}} = \sqrt{\frac{{(2 \times 1.2 \sin 25)}^{2} (0.015) (9.8) \tan 25}{9 \times 10^{9}}} = 2.8 \times 10^{- 6} C$

Therefore, the magnitude of q is $2.8 \times 10^{- 6} C$

Direction

From step 2, we get

${(\sin θ)}^{2} \tan θ = \frac{q^{2} k}{4 L^{2} m g} = 0.33$

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

{(\sin 39.5)}^{2} \tan 39.5 \approx 0.333

Therefore, 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°.