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 uniform vertical electric field E . The value of E is

Question

​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 uniform vertical electric field E . The value of E is <img class="img-fluid question-image" alt="Question" src="https://cdn.tardigrade.in/q/nta/p-pqwt73dqds7d.png" /> (A) (m g/q) (B) (m g/2 q) (C) (m g textittan θ /2 q) (D) none

Tardigrade Admin · Accepted Answer

<img class="img-fluid question-image" alt="Solution" src="https://cdn.tardigrade.in/q/nta/p-pqwt73dq23h2.png" /> For rotational equilibrium, displaystyle ∑ τP = 0 ⇒ mgacosα -qE(.2acos .) = 0 ⇒ E = (m g/2 q)

Q. 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 uniform vertical electric field $E$ . The value of $E$ is

$\frac{m g}{q}$

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

$\frac{m g tan θ}{2 q}$

none

For rotational equilibrium, $\sum τ_{P} = 0$
$\Rightarrow m g a cos α - qE (2 a cos) = 0$
$\Rightarrow$ $E = \frac{m g}{2 q}$

Q. ​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 uniform vertical electric field E . The value of E is

qmg​

2qmg​

2qmgtanθ​