A charged ball B hangs from a silk thread S,w | vedclass

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(B) Let $T$ be the tension in the silk thread,$q$ be the charge on the ball,$m$ be the mass of the ball,and $E$ be the electric field produced by the

Vedclass · Accepted Answer

$	an 	heta$

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(B) Let $T$ be the tension in the silk thread,$q$ be the charge on the ball,$m$ be the mass of the ball,and $E$ be the electric field produced by the large charged sheet.At equilibrium,the forces acting on the ball are:$1$. Tension $T$ along the thread.$2$. Gravitational force $mg$ acting downwards.$3$. Electric force $qE$ acting horizontally away from the sheet.Resolving the tension $T$ into components:$T \sin 	heta = qE$ (horizontal component)$T \cos 	heta = mg$ (vertical component)Dividing the two equations:$\frac{T \sin 	heta}{T \cos 	heta} = \frac{qE}{mg}$$	an 	heta = \frac{qE}{mg}$The electric field $E$ due to a large charged conducting sheet is given by $E = \frac{\sigma}{\varepsilon_0}$.Substituting this into the equation:$	an 	heta = \frac{q}{mg} \left( \frac{\sigma}{\varepsilon_0} ight)$Since $q, m, g,$ and $\varepsilon_0$ are constants,we have:$	an 	heta \propto \sigma$ or $\sigma \propto 	an 	heta$.Therefore,the correct option is $(b)$.

$A$ 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

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