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

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$\frac{m g}{q}$

$\frac{m g \tan θ}{q}$

$\frac{m g}{2 q}$

$\frac{m g \tan θ}{2 q}$

answer is B.

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Detailed Solution

$\begin{array}{l} N = m g \cos θ - - - - (1) \\ f_{s} = μ_{s} N = m g \sin θ - - - - (2) \\ τ_{c m} = 0 \\ 2 q E \sin θ r = f_{s} r = m g \sin θ r \\ 2 q E = m g \\ E = \frac{m g}{2 q} \end{array}$

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