Question

A small sphere with mass m carries a positive charge q and is attached to one end of a silk fiber of length L. The other end of the fiber is attached to a large vertical insulating sheet that has a positive surface charge density s. Show that when the sphere

is in equilibrium, the fiber makes an angle equal to $\mathbf{arctan}\left(\frac{\mathrm{\sigma q}}{2_{\mathrm{\epsilon o}}\mathrm{mg}}\right)$with the vertical sheet.

Short answer

Proven below

Step-by-step solution

Step 1:

Required: Proven that$\mathrm{\theta }=\arctan \left(\frac{\mathrm{\sigma q}}{2_{\mathrm{\epsilon o}}\mathrm{mg}}\right)$

Figure

Calculation

The electric field of the infinite sheet is given by

$\mathrm{E}=\frac{\mathrm{\sigma }}{2_{\mathrm{\epsilon o}}}$ (1)

So the force acting on the ball is given by

$\mathrm{F}=\mathrm{E}=\frac{\mathrm{\sigma q}}{2_{\mathrm{\epsilon o}}}$ (2)

The electric force is equal to the tension component

$\mathrm{T}\sin \left(\mathrm{\theta }\right)=\frac{\mathrm{\sigma q}}{2_{\mathrm{\epsilon o}}}$ (3)

And the vertical component is equal to the weight

$\mathrm{T}\sin \left(\mathrm{\theta }\right)=\mathrm{mg}$ (4)

Dividing (3) on (4) to get

$$\begin{gathered} \tan (\mathrm{\theta })=\frac{\mathrm{\sigma q}}{2{\mathrm{mg}}_{\mathrm{\epsilon o}}} \\ \Rightarrow \mathrm{\theta }=\arctan \frac{\mathrm{\sigma q}}{2_{\mathrm{\epsilon o}}\mathrm{mg}} \end{gathered}$$