Q63.P - University Physics with Modern Physics

Question

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.)

                                                                   

Step-by-Step Solution

Answer

(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 1: Forces on each ball

The diagram shows forces on each ball

Step 2: 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 

 $T\cos \theta =mg$

From the above equations get

 $F=mg\tan \theta$

The repulsive force F is given by 

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

Step 3: Substitution

From step 2, we get 

 $$\begin{gathered} q=\sqrt{\frac{{\left(2L\sin \theta \right)}^{2}mg\tan \theta }{K}} \\ =\sqrt{\frac{{\left(2\times 1.2\sin 25\right)}^2\left(0.015\right)\left(9.8\right)\tan 25}{9\times {10}^9}} \\ =2.8\times {10}^{-6}C \end{gathered}$$

 

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

Step 4: Direction

From step 2, we get

 ${\left(\sin \theta \right)}^2\tan \theta =\frac{q^{2}k}{4L^{2}mg}=0.33$

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

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

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°.