Physics, 06.05.2020 05:02 elf55

Two 3 kg spherical masses are connected by a 20 m long thread of negligible mass that does not stretch. The masses are charged so they have the same amount of negative charge and the system is hung from the ceiling by the center of the thread.

Determine the charge of each sphere if the angle between the suspended sides of the thread is 17 degree.

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Answer from: ddddre4460

The charge of each sphere is $q = - 1.84 *10^{-4}C$

Explanation:

The free body diagram of this question is shown on the first uploaded image

From the question we are told that

The mass of the two sphere is $m = 3 \ kg$

The length of the connection is $l = 20\ m$

The angle between the suspended side is $\theta = 17 ^o$

At Equilibrium the force acting at the horizontal is = 0N and the net force acting at the vertical is zero

This can be represented mathematically as

$\sum F_y = 0N , \ \ and \ \ \sum F_x = 0N$

For $\sum F_y = 0N$

$T cos \theta = mg$

=> $cos \theta = \frac{mg}{T}$

For $\sum F_x = 0N$

$T sin \theta = F$

=> $sin \theta = \frac{F}{T}$

Now $tan \theta = \frac{sin \theta }{cos \theta }$

$= \frac{\frac{F}{T} }{\frac{mg}{T} }$

$= \frac{F}{mg}$

Where F is the electrical force which is mathematically represented as

$F = \frac{kq^2}{r^2}$

Therefore

$tan \theta = \frac{kq^2}{mgr^2}$

Where is the distance between the two masses and from the diagram it is

$r = 2l sin \theta$

$tan \theta = \frac{kq^2 }{(2l sin \theta )^2 * mg}$

making q the subject of the formula

$q = \sqrt{\frac{mg}{k} * tan \theta (2l sin \theta )^2 }$

Where k is the Coulomb's constant with a value of $k = 9*10^{9} kg \ cdot m^3 s^{-4} A^{-2}$

Substituting values

$q = \sqrt{\frac{3 * 9.8 }{9*10^9} * tan (17) (20 sin 17)^2 }$

$q = - 1.84 *10^{-4}C$

The negative sign is because we are told from the question that they are negatively charged

Two 3 kg spherical masses are connected by a 20 m long thread of negligible mass that does not stret

Answer from: Quest

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Answer from: Quest

5.5 n====== $\sum f = ma$ assuming that $\sum f = f_{app}$ we have m = 0.145 kg a = 38 m/s²therefore $f_{app} = (0.145\text{ kg})(38\text{ m/s}^2) = 5.51\text{ n}$ 5.51 n, but two significant figures.final answer: 5.5 n

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