Question

Question 6 In the figure a small nonconducting ball of mass m 0.98 g and charge q = 2.8 10 C distribute uniformy through its volume na s rom an insulting read that makes an angle θ 390 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. m, g Number Units the tolerance is +/-596

Answer 1

6.

Given : m = 0.98×10^-3 g , q = 2.8×10^-8 C, $\theta$=39°

Solution:

Let the tension in the thread be T. Assuming the ball is not moving , the net force on the ball is zero.

qE = Tsin$\theta$

mg = Tcos$\theta$

$T = \frac{mg}{cos\theta}$

$T = \frac{(0.98*10^-6 kg)(9.81 m/s^2)}{cos(39�)}$

T = 1.24×10^-5 N

Solving for electric field at the location of ball:

$E = \frac{Tsin\theta}{q}$

$E = \frac{(1.24*10^-5)sin(39�)}{2.8*10^-8}$

E = 278.7 N/C

We know that, electric field due to infinite sheet of charge is given by: E = 2$\pi$k$\sigma$

Therefore , surface charge density is given by:

$\sigma = \frac{E}{2\pi k}$

$\sigma = \frac{278.7 N/C}{2\pi (9*10^9 )}$

$\sigma$ = 4.93×10^-9 C/m² or say 4.93 nC/m²