A small bob of mass 100 mg and charge +10 m | vedclass

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Question

(D) At equilibrium,the forces acting on the bob are the tension $T$ in the string,the gravitational force $mg$ acting downwards,and the electric force

Vedclass · Accepted Answer

$1.77$

Vedclass · Answer

(D) At equilibrium,the forces acting on the bob are the tension $T$ in the string,the gravitational force $mg$ acting downwards,and the electric force $F_e = qE$ acting horizontally away from the sheet.For an infinitely long nonconducting sheet,the electric field is $E = \frac{\sigma}{2\varepsilon_0}$.Resolving the forces at equilibrium:$T \sin(45^{\circ}) = F_e = qE = q \left( \frac{\sigma}{2\varepsilon_0} ight)$$T \cos(45^{\circ}) = mg$Dividing the two equations:$	an(45^{\circ}) = \frac{q\sigma}{2\varepsilon_0 mg}$Since $	an(45^{\circ}) = 1$,we have:$1 = \frac{q\sigma}{2\varepsilon_0 mg} \implies \sigma = \frac{2\varepsilon_0 mg}{q}$Given values: $m = 100 \ mg = 100 	imes 10^{-6} \ kg = 10^{-4} \ kg$,$q = 10 \ \mu C = 10^{-5} \ C$,$g = 10 \ m/s^2$,$\varepsilon_0 = 8.85 	imes 10^{-12} \ F/m$.Substituting the values:$\sigma = \frac{2 	imes 8.85 	imes 10^{-12} 	imes 10^{-4} 	imes 10}{10^{-5}}$$\sigma = 17.7 	imes 10^{-12} 	imes 10^2 = 17.7 	imes 10^{-10} \ C/m^2 = 1.77 	imes 10^{-9} \ C/m^2 = 1.77 \ nC/m^2$.

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