Two identical 9.60-g metal spheres (small enough to be treated as particles) are hung from separate 500-mm 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?

1.349 × 10¹² electrons

Explanation:

At equilibrium, the tension T in the string is resolved into horizontal and vertical components. Tcos17 being the vertical component and Tsin17 the horizontal component. The electrostatic force of repulsion at equilibrium, F = kq²/r² where q = excess charge on sphere and r = distance apart at equilibrium,F acts horizontally to the left and its weight, mg acts vertically downwards.

For equilibrium, sum of horizontal components = 0 and  sum of horizontal components = 0. So, Tsin17 - F = 0

Tsin17 = F ⇒ Tsin17 = kq²/r² (1)

Also Tcos17 - mg = 0  ⇒ Tcos17 = mg (2)

Dividing (1) by (2), we have

Tsin17/Tcos17 = kq²/r² ÷ mg

tan17 = kq²/mgr²

q = r√(mgtan17/k)

We find r using cosine rule r = √(500² + 500² -2 × 500²cos 2 × 17) since the string and the masses form an isosceles triangle at equilibrium.

r = √(2 × 500² -2 × 500²cos34) = 500√2(1 - cos 34) = 500√2 × 0.1710 =120.89 mm = 0.1209 m

substituting m = 9.60 g = 9.6 × 10⁻³ g, k = 9 × 10⁹ Nm²/C² and r into q, we have,

q = r√(mgtan17/k)

  = 0.1209 m√(9.6 × 10⁻³ g × 9.8 m/s² × tan17/9 × 10⁹ Nm²/C²)

  = 0.1209 m√(28.76 × 10⁻³ N/9 × 10⁹ Nm²/C²)

  =  0.1209 m√(3.195 × 10⁻¹² C²/m²)

  =  0.1209 m × 1.7877 × 10⁻⁶ C/m

  = 0.2161 × 10⁻⁶ C

  = 2.161 × 10⁻⁷ C

To find the number of surplus electrons, n on each sphere, we divide q by e the electron charge.

So, n = q/e = 2.161 × 10⁻⁷ C ÷ 1.602 × 10⁻¹⁹ C = 0.6825 × 10¹² = 1.349 × 10¹² electrons

0.3x0.21x0.08kg=mass 1 sheet

mass 500 = 500xmass 1

volume pile = 0.3x0.2x0.048

density=mass/volume= 500xmass 1/0.3x0.2x0.048

v patient student

heat the gas and measure the pressure