Question

A point charge of mass m and charge q is rigidly attached to a non conducting block of mass M which is placed on rough surface with friction coefficient $\mu$ as shown in figure. Minimum magnitude of electric field required so that the block starts moving :

• A. $\frac{\mu (m+M)g}{q}$
• B. $\frac{\mu (m+M)g}{q\sqrt{1+{\mu }^2}}$
• C. $\frac{\mu \cdot Mg}{q}$
• D. $\frac{\mu \cdot Mg}{q\sqrt{1+{\mu }^2}}$

Answer 1

Answer: (B)

$\frac{\mu (m+M)g}{q\sqrt{1+{\mu }^2}}$

Equating the forces in vertical direction:
$N+qEsin\theta =(M+m)g$
$N=(M+m)g-qEsin\theta$ ...(i)
Equating the forces in horizontal direction:
$f=qEcos\theta$ ...(ii)
As the block is just about to move, maximum friction $f_{max}=\mu N$ will be acting. Substituting value of f in equation (ii) from equation (i)
$\therefore \mu \left[\right(M+m)g-qEsin\theta ]=qEcos\theta$ ...(iii)
$\therefore \mu (M+m)g=qE(cos\theta +\mu sin\theta )$ ...(iv)

Now for E to be minimum $\frac{dE}{d\theta }$ to be zero. differentiating equation (iv)
$\frac{d\left(\mu \right(M+m\left)g\right)}{d\theta }=\frac{d\left(qE\right(cos\theta +\mu sin\theta \left)\right)}{d\theta }$
$0=q\frac{dE}{d\theta }(cos\theta +\mu sin\theta )+qE(-sin\theta +\mu cos\theta )$...(v)
substituting $\frac{dE}{d\theta }=0$ in equation (v)
$(-sin\theta +\mu cos\theta )=0$
$\therefore \mu =tan\theta$
Substituting value of $\mu$ in equation (iv)
$\therefore cos\theta +\mu sin\theta =cos\theta +\frac{si{n}^2\theta }{cos\theta }$
$=\frac{1}{cos\theta }$
$=sec\theta$
$\therefore \mu (M+m)g=qEsec\theta$
$\therefore E=\frac{\mu (M+m)g}{qEsec\theta }$
Again,
$sec\theta =\sqrt{1+ta{n}^2\theta }$
$=\sqrt{1+{\mu }^2}$ $\therefore E=\frac{\mu (M+m)g}{q\sqrt{1+{\mu }^2}}$