## kanwal32 one year ago 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:

1. kanwal32

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2. kanwal32

@IrishBoy123

3. kanwal32

@ParthKohli

4. kanwal32

@Luigi0210

5. kanwal32

@Miracrown

6. anonymous

Hint: the value of E should hold the gravity of the wheel.

7. Michele_Laino

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

8. Michele_Laino

so, for equilibrium we can write these two equations: $\Large \left\{ \begin{gathered} M{\mathbf{g}} + q{\mathbf{E}} + \left( { - q{\mathbf{E}}} \right) + {\mathbf{R}} = {\mathbf{0}} \hfill \\ {\mathbf{OP}} \times q{\mathbf{E}} + {\mathbf{OQ}} \times \left( { - q{\mathbf{E}}} \right) + {\mathbf{OQ}} \times {\mathbf{R}} = {\mathbf{0}} \hfill \\ \end{gathered} \right.$

9. Michele_Laino

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

10. Michele_Laino

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: $\Large - {R_x} \leqslant {\mu _S}{R_z}$

11. IrishBoy123

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