a What equal positive charges would have to be placed on two

(a) What equal positive charges would have to be placed on two celestial objects with masses 2.72 × 10²⁵ and 9.77 × 10²⁵ kg to neutralize their gravitational attraction? (b) What mass of hydrogen would be needed to provide the positive charge calculated in (a)?

Solution

1 a] F _{electrostatic}= F _{gravitational}

Kq²/r²= Gm₁m₂/r²

9*10⁹*q²=6.67*10^-11* 2.72 × 10²⁵ *9.77 × 10²⁵

q²=6.67*10^-11* 2.72 × 10²⁵ *9.77 × 10²⁵ / 9*10⁹ = 1.97*10⁴⁰

q= 1.40*10²⁰ C

1 b]

q= ne

n= q/e= 1.40*10²⁰/1.6*10^-19= 0.877*10³⁹ -----------------------is the total number of protons needed

Mass of hydrogen needed= 0.877*10³⁹ * 1.67*10^-27 kg = 1.46*10¹² kg

2] Given:-

In the figure, sin = ½ x/L = ½ (6.4)/160 =0.02

Let T is cable tension force,

F_horizontal = 0

T sin - kq²/x² = 0

T sin = kq²/x² . . . . . . . . . . . . . (eqs 1)


F_vertical = 0

T cos - mg = 0

T cos = mg. . . . . . . . . . . . . (eqs 2)


equation 1 divided by equation 2,

tan = kq²/(mgx²)

Assume that is so small that tan can be replaced by its approximate equal, sin

0.02 = (9 * 10^9)q²/(9.3*10^-3 * 9.8 * 0.064²)

0.02= q²*2.4*10¹³

q²= 0.02/2.4*10¹³ =8.29*10^-16
q = 2.88 * 10^-8Coulomb

3] a]

F _{net on particle1}= F₁₂+F₁₃+F₁₄

=Kq₁q₂/d²+kq₁q₃/ (2d)²+kq₁q₄/ (3d)²

= kq₁[q₂/d²+q₃/ (2d)²+q₄/ (3d)² ]

F _net-1= 9*10⁹*4e[-e/d²+e/4d²+12e/9d²] =9 *10⁹*4e* e [-1+0.25+1.33] / d²

F _net-1 = 21*10⁹*e²/d²

3]b]

F _{net on particle2}= F₂₁+F₂₃+F₂₄

=Kq₂q₁/d²+kq₂q₃/ d²+kq₂q₄/ (2d)²

= kq₂[q₁/d²+q₃/ d²+q₄/ (2d)² ]

F _net-1= 9*10⁹*-e[4e/d²+e/d²+12e/4d²] =9 *10⁹*-e* e [4+1+3] / d²

F _net-1 = -72*10⁹*e²/d²