Answer:
the magnitude of the charge Q on each plate is [tex]3.053 *10^{-8} \ C[/tex]
Explanation:
Given that :
mass (m) = [tex]7.00 *10 ^{-3} \ kg[/tex]
charge (q) = +0.155 µC = [tex]+0.155 *10^{-6}\ C[/tex]
angle [tex]\theta = 30^0 \ C[/tex]
Area A on each plate = 0.0135 m²
From the diagram below;
[tex]tan \ \theta = \frac{Eq}{mg}[/tex] ----- equation (1)
Also by using Gauss Law ;
[tex]Q = \epsilon_0 \phi[/tex]
[tex]Q = \epsilon_0EA[/tex] ----- equation (2)
Combination equation 1 and 2 together ; we have
[tex]Q = \frac{\epsilon_0\ * \ m\ *\ g \ \ * \ tan \theta \ * \ A}{q}[/tex]
[tex]Q = \frac{(8.85*10^{-12}C^2/N.m^2 )\ * \ (7.00*10^{-3} kg)\ *\ (9.8 m/s^2) \ \ * \ tan \(30 \ * \ (0.0135 m^2)}{0.155*10^{-6}\ C}[/tex]
[tex]Q = 3.053 *10^{-8} \ C[/tex]
