A small bob of mass 100 mg and charge +10 μ C is connected to an insulating string of length 1 m . It is brought near to an infinitely long non-conducting sheet of charge density ' σ ' as shown in figure. If string subtends an angle of 45° with the sheet at equilibrium the charge density of sheet will be.
(Given, ε₀=8.85 × 10⁻¹² F/m and acceleration due to gravity, g=10 m/s² )

Solution

Physics sir · 4 steps

At 45° the electric push and the weight are equal, so set qE = mg and the string length drops out.

1. Forces on the bob. Three forces act on the bob: the weight mg downward, the tension T along the string, and the electric force qE horizontally away from the positive sheet. Resolve T along the vertical and the horizontal.
2. Balance them. Vertically T cos 45° = mg and horizontally T sin 45° = qE. Divide the two equations to get tan 45° = qE/mg. Since tan 45° = 1, qE = mg, and the 1 m length plays no part.

   q σ/2ε₀ = mg ⇒ σ = (2ε₀ mg)/q

3. Put in numbers. m = 100 mg = 10⁻⁴ kg, so mg = 10⁻³ N, and q = 10⁻⁵ C. Then σ = (2 × 8.85 × 10⁻¹² × 10⁻³)/10⁻⁵ = 1.77 × 10⁻⁹ C m⁻².
4. So. The charge density of the sheet is 1.77 nC m⁻², which is option (a).

The idea to keep. For a charged bob hanging near a sheet, tan θ = qE/mg. The field of an infinite non-conducting sheet is σ/2ε₀; keep σ/ε₀ for the outside of a conductor.

Answer from the NTA final answer key. Solution by PrepWiser.