Answer:
The angle is 18.3 degree.
Explanation:
A uniformly charged infinite plane, density σ = 4 x 10^-9 C/cm^2, is placed vertically in air. A small ball of mass 8 g, with charge q = 10^-8 C, hangs close to the plane, so that the string is initially parallel to the plane. Take g = 9.8m/s2. When in equilibrium, by what angle is the string hanging the ball to the plane?
surface charge density, σ = 4 x 10^-5 C/m^2
Charge, q = 10^-8 C
mass, m = 0.008 kg
Let the angle is A and the tension in the string is T.
The electric field due to a plane is
[tex]E =\frac{\varepsilon \sigma }{2\varepsilon o}\\\\E =\frac{4\times 10^{-5}}{2\times 8.85\times 10^{-12}}\\\\E = 2.26\times 10^6 V/m \\[/tex]
Now equate the forces,
[tex]T sin A = q E.... (1)\\\\T cos A = m g ..... (2)\\\\divide (1) by (2)\\\\tan A = \frac{10^{-8}\times 2.6\times 10^6}{0.008\times 9.8}\\\\tan A = 0.33\\\\ A = 18.3 degree[/tex]
