Question

One end of a 10 cm 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 of mass 10 g and a charge of $4·0\times {10}^{-6}\mathrm{C}$. In equilibrium, the thread makes an angle of 60° with the vertical. Find the surface charge density on the plate.

Answer 1

There are two forces acting on the ball. These are
(1) Weight of the ball, W = mg
(2) Coulomb force acting on the charged ball due to the electric field of the plate, F = qE
Due to these forces,a tension develops in the thread.

Let the surface charge density on the plate be σ.

Electric field of a plate,
E = $\frac{\sigma }{2{\in }_0}$
It is given that in equilibrium, the thread makes an angle of 60° with the vertical.

Resolving the tension in the string along horizontal and vertical directions, we get:

$$\begin{gathered} T\cos 60°=mg \\ T\sin 60°=\mathit{}qE \\ \Rightarrow \tan 60°=\frac{\mathit{q}\mathit{E}}{\mathit{m}\mathit{g}} \\ \Rightarrow E=\frac{m\mathrm{g}\tan 60°}{q} \end{gathered}$$
Also, electric field due to a plate,
$$\begin{gathered} E=\frac{\sigma }{2{\in }_0}=\frac{mg\tan {60}^{\mathrm{o}}}{q} \\ \sigma =\frac{2{\in }_{0}m\mathrm{g}\tan 60°}{q} \\ \sigma =\frac{2\times \left(8.85\times {10}^{-12}\right)\times \left(10\times {10}^{-3}\times 9.8\right)\times 1.732}{4.0\times {10}^{-6}} \\ \sigma =7.5\times {10}^{-7}\mathrm{C}/{\mathrm{m}}^2 \end{gathered}$$