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 na

Question

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?

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Kim Chi 4 years 2021-07-24T02:25:53+00:00 2 Answers 41 views 0

Answers ( )

    0
    2021-07-24T02:27:16+00:00

    Answer:

    n = 3.94 *10^{12}

    Explanation:

    Using the expression of electric force:

    F = \frac{kq_1q_2}{r^2}

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

    Then;

    mg tan \theta = \frac{kqq}{d^2}

    making a the subject of the formula ; we have:

    q^2 = \frac{d^2 mg tan \theta}{k}

    q= \sqrt{ \frac{d^2 mg tan \theta}{k}}    ————– equation (1)

    By trigonometric rule:

    sin \theta = \frac{opposite}{hypotenuse}

    let \frac{d}{2} be the opposite side; and

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

    Then

    sin \theta = \frac{d}{2l}

    d = 2 \ l sin \theta

    Replacing d = 2 \ l sin \theta into equation  1 ; we have:

    q =  2\ l sin \theta \sqrt{\frac{mgtan \theta}{k}}

    Equation for number of charge particle n = \frac{q}{e}

    So;

    n = \frac{2 \ l sin\theta}{e} \sqrt{\frac{mgtan \theta}{k}}

    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;

    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} }

    n = 3.94 *10^{12}

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

    0
    2021-07-24T02:27:39+00:00

    Answer:

    The number of electrons are 3.94×10¹²

    Explanation:

    The translational equilibrium is:

    Tsin\theta -F_{21} =0\\Tsin\theta =F_{21}

    Dividing the expression by Tcosθ = mg

    \frac{Tsin\theta}{Tcos\theta } =\frac{F_{21} }{mg} \\tan\theta =\frac{F_{21} }{mg}\\F_{21} =mg tan\theta

    The equation for electric force is:

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

    if:

    d=2Lsin\theta

    n=\frac{q}{\epsilon }

    Replacing:

    n=\frac{2Lsin\theta }{\epsilon } \sqrt{\frac{mgtan\theta }{k} }

    Where

    L = 700 mm = 0.7 m

    ε = 1.6×10⁻¹⁹C

    m = 7.1 g = 0.0071 kg

    k = 8.99×10⁹N m²/C²

    Replacing:

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

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