Question

Two small balls, each of mass 5.0 g, are attached to silk threads 50 cm long, which are in turn tied to the same point on the ceiling, as shown below. When the balls are given the same charge Q, the threads hang at 5.0°5.0° to the vertical, as shown below.

Answer 1

Given

Mass of the balls m = 5 g = 5.00 x 10^-3 kg

Length of the pendulum L = 50 cm = 0.50 m

Angle θ = 5.0^o

Solution

Since the balls are at equilibrium the net force acting on them will be zero.

The forces acting on the in y direction

W = Tcosθ

mg = Tcosθ

T = mg/cosθ

The forces acting on the x direction

F = Tsinθ

F = (mg/cosθ) x sinθ

F = mgtanθ

kQ²/r² = mgtanθ

Q² = mgtanθ . r²/k

Since r = 2Lsinθ

Q² = mgtanθ.(2Lsinθ)²/k

Q = √(mg.tanθ.(2Lsinθ)²/k)

Q = (2L.sinθ) x √(mg.tanθ/k)

Q = 2 x 0.50 x sin5^o√(5.00 x 10^-3 x 9.8 x tan5^o / 9 x 10⁹)

Q = sin5^o √( 5.44 x 10^-12 x tan5^o)

Q = 60.2 x 10^-9 C