Question
A charged ball $$B$$ hangs from a silk thread $$S,$$ which makes an angle $$\theta $$ with a large charged conducting sheet $$P,$$ as shown in the figure. The surface charge density $$\sigma $$ of the sheet is proportional to
A charged ball $$B$$ hangs from a silk thread $$S,$$ which makes an angle $$\theta $$ with a large charged conducting sheet $$P,$$ as shown in the figure. The surface charge density $$\sigma $$ of the sheet is proportional to
A.
$$\cot \theta $$
B.
$$\cos \theta $$
C.
$$\tan \theta $$
D.
$$\sin \theta $$
Answer :
$$\tan \theta $$
Solution :
$$\eqalign{ & T\sin \theta = \frac{\sigma }{{{\varepsilon _0}K}}.q\,......\left( {\text{i}} \right) \cr & T\cos \theta = mg\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) by (ii),
$$\eqalign{ & \tan \theta = \frac{{\sigma q}}{{{\varepsilon _0}K.mg}}\,\,\,\, \cr & \therefore \sigma \, \propto \tan \theta \cr} $$
$$\eqalign{ & T\sin \theta = \frac{\sigma }{{{\varepsilon _0}K}}.q\,......\left( {\text{i}} \right) \cr & T\cos \theta = mg\,......\left( {{\text{ii}}} \right) \cr} $$
Dividing (i) by (ii),
$$\eqalign{ & \tan \theta = \frac{{\sigma q}}{{{\varepsilon _0}K.mg}}\,\,\,\, \cr & \therefore \sigma \, \propto \tan \theta \cr} $$
