Question

In the figure a small, nonconducting ball of mass m = 1.1 mg and charge q = 1.8 × 10^-8 C (distributed uniformly through its volume) hangs from an insulating thread that makes an angle θ = 45° with a vertical, uniformly charged nonconducting sheet (shown in cross section). Considering the gravitational force on the ball and assuming the sheet extends far vertically and into and out of the page, calculate the surface charge density σ of the sheet.

Answer 1

The non-conducting ball hanging through the sheet is represented as follows: T cose E T sine mg Figure a: A non conducting ball on the vertical sheet. Here, T is the tension, m is the mass, o is the angle at which ball is suspended, g is the gravitational acceleration, q is the charge, and E is the electric field. The equation in vertical direction is represented as follows: T cos 0 = mg mg cose T = The equation in the horizontal direction is represented as follows: I sin 0 =qE TqE sin Ꮎ

Substitute mg for T. cose mgqE cos o sin e E = mg tan [1mg ||(9.8 m/s²) 1.1 mg 100 kg (9.8 m/s) (1.8 x 10-C) — tan (45°) = 598.99 N/C The expression for the electric field of the non-conducting sheet is represented as follows: 28. o is surface charge density and a = 8.85x10-12 CPN-mis the permittivity of free space. Therefore, o=2EE. = 2 (598.99 N/C)(8.85x10-12 C°/Nm) = 1.06x10-6 C/m Therefore, the surface charge density of sheet is 1.06 x 10° C/mºl