Question

A wheel having mass $1\text{kg}$ has charges $10\text{n}\text{C}$ and $-10\text{n}\text{C}$ on diametrically opposite points as shown in figure. It remains in equilibrium on a rough inclined plane in the presence of a uniform vertical electric field $E$. What is the value of $E$ ?
(Take $g=10{\text{m/s}}^2$)

• A. ${10}^9\text{N/C}$
• B. $5\times {10}^8\text{N/C}$
• C. $5\sqrt{3}\times {10}^8\text{N/C}$
• D. $8\times {10}^8\text{N/C}$

Answer 1

The correct option is B $5\times {10}^8\text{N/C}$

Given:
$q_1=10\text{nC}=10\times {10}^{-9}\text{C};m=1\text{kg}$

$q_2=-10\text{nC}=-10\times {10}^{-9}\text{C}$



Since, the direction of electric field is vertically upward.
So, point $\text{B}$ experience upward electric force $F_1$ and point $\text{A}$ experience download electric force $F_2$.

$|F_1|=|F_2|=qE=10\times {10}^{-9}E.....\left(1\right)$

and, $mg=1\times 10=10\text{N}$

We know that for equilibrium

${\tau }_{net}=0$

on taking ${\tau }_{net}$ about point $\text{A}$,

${\tau }_A=0$

$\Rightarrow r\left(mg\sin {60}^{\circ }\right)-2r\left(F_1\sin {60}^{\circ }\right)=0$

$\Rightarrow r\left(mg\sin {60}^{\circ }\right)=2r\left(F_1\sin {60}^{\circ }\right)$

$\Rightarrow F_1=\frac{mg}{2}=5$

from eq. (1)

$\Rightarrow 10\times {10}^{-9}\times E=5$

$\Rightarrow E=5\times {10}^8\text{N/C}$

Hence, option $\left(b\right)$ is the correct answer.

$\begin{array}{c}\text{Why this question}?\\ \text{Concept - (1) For equilibrium of a body,}{F}_{net}=0\\ \text{and}{\tau }_{net}=0\\ \text{(2) Torque due to force passing through axis}\\ \text{of rotation is zero because}{r}_{\perp }=0\end{array}$