Question

One end of a 10cm long silk thread is fixed to a large vertical surface of a charged non conducting plate and the other end is fastened to a small ball having a mass of $10\,g$ and charge of $4.0 \times {10^{ - 6}}\,C$. In equilibrium, the thread makes an angle of ${60^ \circ }$ with the vertical. Find the surface charge density on the plate.

Answer 1

Hint: The surface charge density of the plate can be determined by creating the tension equation of the rope and the torque equation of the charge. By dividing these two equations, the surface charge density on the plate can be determined.

Complete step by step solution:
Given that,
The length of the thread is, $L = 20\,cm$,
The mass of the sphere is, $m = 10\,g$,
The charge of the plate is, $q = 4.0 \times {10^{ - 6}}\,C$.
The angle of the thread is, $\theta = {60^ \circ }$.
Now, the tension in the thread is given by,
$T\cos \theta = mg\,................\left( 1 \right)$
Where, $T$ is the tension of the thread, $m$ is the mass tied on the thread and $g$ is the acceleration due to gravity.
Now, the torque is given by,
$T\sin \theta = qE\,...............\left( 2 \right)$
Where, $T$ is the torque, $q$ is the charge and $E$ is the electric dipole.
By dividing the equation (2) by the equation (1), then
$\dfrac{{T\sin \theta }}{{T\cos \theta }} = \dfrac{{qE}}{{mg}}$
By cancelling the same terms in the above equation, then the above equation is written as,
$\dfrac{{\sin \theta }}{{\cos \theta }} = \dfrac{{qE}}{{mg}}$
The term $\dfrac{{\sin \theta }}{{\cos \theta }} = \tan \theta $, then the above equation is written as,
$\tan \theta = \dfrac{{qE}}{{mg}}$
By rearranging the terms in the above equation, then
$\dfrac{{\tan \theta \times mg}}{q} = E$
The electric dipole is also written as,
$\dfrac{{\tan \theta \times mg}}{q} = \dfrac{\sigma }{{2{\varepsilon _0}}}$
By rearranging the terms in the above equation, then the above equation is written as,
$\dfrac{{\tan \theta \times mg \times 2{\varepsilon _0}}}{q} = \sigma $
By substituting the known values in the above equation, then
$\sigma = \dfrac{{\tan {{60}^ \circ } \times 10 \times {{10}^{ - 3}} \times 9.8 \times 2 \times 8.8 \times {{10}^{ - 12}}}}{{4.0 \times {{10}^{ - 6}}}}$
The value of the $\tan {60^ \circ } = 1.732$, then
$\sigma = \dfrac{{1.732 \times 10 \times {{10}^{ - 3}} \times 9.8 \times 2 \times 8.8 \times {{10}^{ - 12}}}}{{4.0 \times {{10}^{ - 6}}}}$
By multiplying the terms in the above equation, then
$\sigma = \dfrac{{2.987 \times {{10}^{ - 12}}}}{{4.0 \times {{10}^{ - 6}}}}$
By dividing the terms in the above equation, then the above equation is written as,
$\sigma = 7.5 \times {10^{ - 7}}\,C{m^{ - 2}}$

Thus, the above equation shows the surface charge density of the plate.

Note: The surface charge density of the plate is directly proportional to the mass of the sphere, acceleration due to gravity and permittivity of the free space and inversely proportional to the charge of the plate. If the charge of the plate increases, the surface charge density decreases.