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Chapter 21: Problem 80

Two balls have the same mass of \(0.681 \mathrm{~kg}\) and identical charges of \(18.0 \mu \mathrm{C} .\) They hang from the ceiling on strings of identical length as shown in the figure. If the angle with respect to the vertical of the strings is \(20.0^{\circ}\), what is the length of the strings?

Short Answer

Expert verified
(where the mass of the balls is \(0.681\mathrm{~kg}\) and the charge of both balls is \(18.0\mu \mathrm{C}\)) Answer: The length of the strings is approximately \(1.01\) meters.

Step by step solution

01

Determine the electric repulsion force

We will use Coulomb's Law to find the force of repulsion between the two charged balls: \(F_e = \cfrac{k|q_1 q_2|}{r^2}\) Where \(F_e\) is the electric force, \(k\) is the electrostatic constant (\(\approx 8.99\times10^9\,\mathrm{N\cdot m^2/C^2}\)), \(q_1\) and \(q_2\) are the charges of the balls, and \(r\) is the distance between them. Since both charges are identical and positive, we have: \(F_e = \cfrac{k(q)^2}{(2L\sin{\theta})^2}\) Where \(q=18.0\mu \mathrm{C}\), \(L\) is the length of the strings, and \(\theta=20.0^{\circ}\).
02

Determine the gravitational force

The gravitational force acting on each ball is given by: \(F_g = mg\) Where \(F_g\) is the gravitational force, \(m=0.681\mathrm{~kg}\) is the mass of the balls, and \(g\approx9.81\mathrm{~m/s^2}\) is the acceleration due to gravity.
03

Find the tension in the strings

The tension in the strings can be found by examining the horizontal and vertical components separately. The horizontal component of the tension (\(T_x\)) equals the electric repulsion force, and the vertical component (\(T_y\)) equals the gravitational force: \(T_x=F_e\) \(T_y=F_g\) Using trigonometry, we can find the total tension in the string as: \(T=\frac{T_y}{\cos\theta}\)
04

Set up the equation to solve for the length of strings (L)

We have the following equation from using the trigonometric and force equations: \(F_e = T\sin\theta\) \(F_g = T\cos\theta\) Divide the two equations to eliminate tension (T): \(\cfrac{F_e}{F_g}=\cfrac{T\sin\theta}{T\cos\theta} \) \(\cfrac{F_e}{F_g}=\tan\theta \) Now we will substitute \(F_e\) and \(F_g\) into the above equation: \(\cfrac{k(q)^2}{(2L\sin{\theta})^2g}=\tan\theta\) Our goal is to solve for \(L\).
05

Solve for the length of the strings (L)

Rearrange the equation to isolate L: \(L=\cfrac{k(q)^2}{4g\sin^2\theta\tan\theta}\) Substitute the given values and constants: \(L=\cfrac{8.99\times10^9 \cdot (18\times10^{-6})^2}{4\times 9.81\times\sin^2(20)\times\tan(20)}\) Now, compute the length: \(L\approx1.01\,\mathrm{m}\) Therefore, the length of the strings is approximately \(1.01\) meters.

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Most popular questions from this chapter

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