If a small sphere of mass m and charge q is h | vedclass

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(A) For the equilibrium of the sphere,the forces acting on it are the tension $T$ in the thread,the gravitational force $mg$ acting downwards,and the

Vedclass · Accepted Answer

$\varepsilon_0\left(\frac{m g}{q}ight) 	an 	heta$

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(A) For the equilibrium of the sphere,the forces acting on it are the tension $T$ in the thread,the gravitational force $mg$ acting downwards,and the electrostatic force $F_e = qE$ acting horizontally away from the plate.Since the plate is a large conducting sheet,the electric field $E$ near its surface is given by $E = \frac{\sigma}{2\varepsilon_0}$. However,for a charged conducting plate,the field is $E = \frac{\sigma}{\varepsilon_0}$.The horizontal component of tension balances the electrostatic force: $T \sin 	heta = qE = q \left(\frac{\sigma}{\varepsilon_0}ight)$.The vertical component of tension balances the gravitational force: $T \cos 	heta = mg$.Dividing the two equations: $\frac{T \sin 	heta}{T \cos 	heta} = \frac{q \sigma / \varepsilon_0}{mg}$.$	an 	heta = \frac{q \sigma}{\varepsilon_0 m g}$.Solving for $\sigma$,we get $\sigma = \frac{\varepsilon_0 m g 	an 	heta}{q}$.

If a small sphere of mass $m$ and charge $q$ is hung from a silk thread at an angle $\theta$ with the surface of a vertical charged conducting plate,then for equilibrium of the sphere,the surface charge density of the plate is

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