A small block of mass 5 , g and charge 5 , mu | vedclass

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(C) For the block to remain at rest on a frictionless inclined plane, the component of gravitational force acting down the plane must be balanced by t

Vedclass · Accepted Answer

$\frac{\sqrt{3}}{2} 	imes 10^4 \, N/C$

Vedclass · Answer

(C) For the block to remain at rest on a frictionless inclined plane, the component of gravitational force acting down the plane must be balanced by the electric force acting up the plane.$mg \sin 	heta = qE$$E = \frac{mg \sin 	heta}{q}$Given: $m = 5 \, g = 5 	imes 10^{-3} \, kg$, $q = 5 \, \mu C = 5 	imes 10^{-6} \, C$, $	heta = 60^{\circ}$, $g = 10 \, m/s^2$.$E = \frac{5 	imes 10^{-3} 	imes 10 	imes \sin(60^{\circ})}{5 	imes 10^{-6}}$$E = \frac{5 	imes 10^{-2} 	imes \frac{\sqrt{3}}{2}}{5 	imes 10^{-6}}$$E = 10^4 	imes \frac{\sqrt{3}}{2} \, N/C$Thus, the magnitude of the electric field is $\frac{\sqrt{3}}{2} 	imes 10^4 \, N/C$.

$A$ small block of mass $5 \, g$ and charge $5 \, \mu C$ is placed on an insulated, frictionless, inclined plane of angle $60^{\circ}$. An electric field is applied parallel to the inclined plane. If the block remains at rest, then the magnitude of the electric field is (Take $g = 10 \, m/s^2$):

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