Question

Two small spheres each of mass 10^-6 kg are suspended from a point by silk threads 50 cm long. tgey are equally charged and repel each other to a distance 20 cm apart. calculate the combined potential at the surface of each shpere.

Answer 1

Answer:

Two small sphere of mass , m =10⁻⁶ Kg are suspended from a point thread of 50cm long as shown in figure. They are equally charged and repel each other to a distance 20cm apart.

Let charge on each sphere is q and tension on string is T

Let thread inclined with vertical at an angle θ

Then, at equilibrium condition ,

Downward force = upward force

mg = Tcosθ

10⁻⁶ × 10 = 10⁻⁵ = Tcosθ ------(1)

Forward force = backward force

Coulombs force = Tsinθ

Kq²/(0.20)² = Tsinθ [ because distance between two sphere is 20cm = 0.2m ]

9 × 10⁹q²/0.04 = Tsinθ

2.25 × 10¹¹ q² = Tsinθ -------(2)

Dividing equation (1) from equation (2)

2.25 × 10¹¹q²/10⁻⁵ = tanθ

Now, tanθ = 10/√{50² - 10²} = 10/20√6 = 1/2√6 [ see figure]

q² = 10⁻¹⁶/{2√6 × 2.25} = 10⁻¹⁶/(2×2.449 × 2.25)

q² = 9.07 × 10⁻¹⁸

taking square root both sides,

q = 3 × 10⁻⁹C