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

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(B) Let $q$ be the charge on the ball $B$ and $m$ be its mass. The electric field $E$ due to the large charged conducting sheet $P$ is given by $E = \

Vedclass · Accepted Answer

$	an 	heta$

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(B) Let $q$ be the charge on the ball $B$ and $m$ be its mass. The electric field $E$ due to the large charged conducting sheet $P$ is given by $E = \frac{\sigma}{2\varepsilon_0}$.The electric force acting on the ball is $F_e = qE = \frac{q\sigma}{2\varepsilon_0}$.At equilibrium,the forces acting on the ball are:$1$. Tension $T$ in the thread.$2$. Gravitational force $mg$ acting downwards.$3$. Electric force $F_e = \frac{q\sigma}{2\varepsilon_0}$ acting horizontally away from the sheet.Resolving the tension $T$ into components:$T \cos 	heta = mg$  $(i)$$T \sin 	heta = F_e = \frac{q\sigma}{2\varepsilon_0}$  $(ii)$Dividing equation $(ii)$ by $(i)$:$\frac{T \sin 	heta}{T \cos 	heta} = \frac{q\sigma / 2\varepsilon_0}{mg}$$	an 	heta = \frac{q\sigma}{2\varepsilon_0 mg}$Since $q, \varepsilon_0, m,$ and $g$ are constants,we have:$\sigma \propto 	an 	heta$.

$A$ 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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