A wheel having mass m has charges +q and -q on diametrically opposite points. It remains equilibrium on a rough inclined plane in the presence of uniform vertical electric field E. The value of E is:

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A wheel having mass m has charges +q and -q on diametrically opposite points. It remains equilibrium on a rough inclined plane in the presence of uniform vertical electric field E. The value of E is:

Physics
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Hint: the value of E should hold the gravity of the wheel.
we have to consider this refernece system |dw:1433059642792:dw| where R is the reaction exerted by the inclined plane on the wheel, and I suppose that the radius of the wheel is r
so, for equilibrium we can write these two equations: \[\Large \left\{ \begin{gathered} M{\mathbf{g}} + q{\mathbf{E}} + \left( { - q{\mathbf{E}}} \right) + {\mathbf{R}} = {\mathbf{0}} \hfill \\ {\mathbf{OP}} \times q{\mathbf{E}} + {\mathbf{OQ}} \times \left( { - q{\mathbf{E}}} \right) + {\mathbf{OQ}} \times {\mathbf{R}} = {\mathbf{0}} \hfill \\ \end{gathered} \right.\]
with the condition: \[\Large {R_x} \leqslant {\mu _S}{R_z}\] with R_x and R_z are the components of R, and \[{\mu _S}\] is the friction coefficient on the plane
developing those vector equation, you should get these scalar equations: \[\Large \left\{ {\begin{array}{*{20}{c}} {Mg\sin \theta + {R_x} = 0} \\ {-Mg\cos \theta + {R_z} = 0} \\ {2qEr\sin \theta + r{R_x} = 0} \end{array}} \right.\] and the additional condition, can be rewritten as follows: \[\Large - {R_x} \leqslant {\mu _S}{R_z}\]
|dw:1433066327238:dw|

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