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(B) The electric field $E$ produced by an infinite nonconducting sheet is given by $E = \frac{\sigma}{2 \varepsilon_0}$.Here,$\sigma = 100 \ \mu C/m^2

Vedclass · Accepted Answer

$1.12$

Vedclass · Answer

(B) The electric field $E$ produced by an infinite nonconducting sheet is given by $E = \frac{\sigma}{2 \varepsilon_0}$.Here,$\sigma = 100 \ \mu C/m^2 = 100 	imes 10^{-6} \ C/m^2$.$E = \frac{100 	imes 10^{-6}}{2 	imes 8.85 	imes 10^{-12}} = \frac{10^{-4}}{17.7 	imes 10^{-12}} = \frac{10^8}{17.7} \ N/C \approx 5.65 	imes 10^6 \ N/C$.The dipole moment $p = q 	imes d = (2 	imes 10^{-6} \ C) 	imes (0.2 \ m) = 0.4 	imes 10^{-6} \ Cm = 4 	imes 10^{-7} \ Cm$.The torque $	au$ acting on the dipole is $	au = pE \sin 	heta$,where $	heta = 30^{\circ}$.$	au = (4 	imes 10^{-7}) 	imes (5.65 	imes 10^6) 	imes \sin(30^{\circ})$$	au = (4 	imes 10^{-7}) 	imes (5.65 	imes 10^6) 	imes 0.5 = 2 	imes 10^{-7} 	imes 5.65 	imes 10^6 = 1.13 \ Nm$.Rounding to the nearest provided option,the correct answer is $1.12 \ Nm$.

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