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wo identical 7.10-gg metal spheres (small enough to be treated as particles) are hung from separate 700-mmmm strings attached to the same nail in a ceiling. Surplus electrons are added to each sphere, and then the spheres are brought in contact with each other and released. Their equilibrium position is such that each string makes a 17.0 ∘∘ angle with the vertical.

How many surplus electrons are on each sphere?

Respuesta :

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Answer:

[tex]n = 3.94 *10^{12}[/tex]

Explanation:

Using the expression of electric force:

[tex]F = \frac{kq_1q_2}{r^2}[/tex]

and replacing mgtanθ for F , q for [tex]q_1 \ and \ q_2[/tex] and d for r in the relation above:

Then;

[tex]mg tan \theta = \frac{kqq}{d^2}[/tex]

making a the subject of the formula ; we have:

[tex]q^2 = \frac{d^2 mg tan \theta}{k}[/tex]

[tex]q= \sqrt{ \frac{d^2 mg tan \theta}{k}}[/tex]    -------------- equation (1)

By trigonometric rule:

[tex]sin \theta = \frac{opposite}{hypotenuse}[/tex]

let [tex]\frac{d}{2}[/tex] be the opposite side; and

l be the adjacent side in the above equation and solve for d:

Then

[tex]sin \theta = \frac{d}{2l}[/tex]

[tex]d = 2 \ l sin \theta[/tex]

Replacing [tex]d = 2 \ l sin \theta[/tex] into equation  1 ; we have:

[tex]q = 2\ l sin \theta \sqrt{\frac{mgtan \theta}{k}}[/tex]

Equation for number of charge particle n = [tex]\frac{q}{e}[/tex]

So;

[tex]n = \frac{2 \ l sin\theta}{e} \sqrt{\frac{mgtan \theta}{k}}[/tex]

Given that :

m = 7.10 g

θ = 17.0 °

g = 9.80 m/s²

k = 8.99× 10 ⁹ N.m ²/C²

l = 700 mm

e = 1.6× 10⁻¹⁰

Then;

[tex]n = \frac{2(700mm(\frac{10^{-3}m}{1mm}) sin17^0 } {1.6*10^{-19}C} \sqrt{\frac{(7.10 g (\frac{10^{-3}kg}{1 g} (9.8 m/s^2)tan 17^0 }{8.99*19^9 N.m^2/C^2} }[/tex]

[tex]n = 3.94 *10^{12}[/tex]

Therefore ; the number of surplus electron that are on each sphere = [tex]3.94 *10^{12}[/tex]

Answer:

The number of electrons are 3.94x10¹²

Explanation:

The translational equilibrium is:

[tex]Tsin\theta -F_{21} =0\\Tsin\theta =F_{21}[/tex]

Dividing the expression by Tcosθ = mg

[tex]\frac{Tsin\theta}{Tcos\theta } =\frac{F_{21} }{mg} \\tan\theta =\frac{F_{21} }{mg}\\F_{21} =mg tan\theta[/tex]

The equation for electric force is:

[tex]F_{21} =\frac{kqq}{d^{2} } \\mgtan\theta =\frac{kqq}{d^{2} }\\q=d\sqrt{\frac{mgtan\theta }{k} }[/tex]

if:

[tex]d=2Lsin\theta[/tex]

[tex]n=\frac{q}{\epsilon }[/tex]

Replacing:

[tex]n=\frac{2Lsin\theta }{\epsilon } \sqrt{\frac{mgtan\theta }{k} }[/tex]

Where

L = 700 mm = 0.7 m

ε = 1.6x10⁻¹⁹C

m = 7.1 g = 0.0071 kg

k = 8.99x10⁹N m²/C²

Replacing:

[tex]n=\frac{2*0.7*sin17}{1.6x10^{-19} } \sqrt{\frac{0.0071*9.8*tan17}{8.99x10^{9} } } =3.94x10^{12}[/tex]