Question

A unit positive point charge of mass $m$ is projected with a velocity $v$ inside the tunnel as shown. The tunnel has been made inside a uniformly charged non-conducting sphere. The minimum velocity with which the point charge should be projected such that it can reach the opposite end of the tunnel is equal to

• A. ${\left[\frac{\rho R^2}{4m{ϵ}_0}\right]}^{1/2}$
• B. ${\left[\frac{\rho R^2}{24m{ϵ}_0}\right]}^{1/2}$
• C. ${\left[\frac{\rho R^2}{6m{ϵ}_0}\right]}^{1/2}$
• D. zero because the initial and the final points are at same potential

Answer 1

The correct option is A ${\left[\frac{\rho R^2}{4m{ϵ}_0}\right]}^{1/2}$

If we throw the charged particle just right of the centre of the tunnel, the particle will cross the tunnel. Hence, applying work energy theorem between start point and centre of tunnel we get

$\Delta K.E=$Work done for half tunnel

$\Rightarrow (\frac{1}{2}m{v}_i^2-\frac{1}{2}m{v}_f^2)=q(V_f-V_i)....\left(1\right)$

Since, the velocity of the charged particle at the centre of tunnel should be zero to just reach at the opposite end .i.e, $v_f=0$

Potential at the left end of the tunnel will be equal to the potential at the surface,
$V_i=\frac{\rho R^2}{3{ϵ}_0}....\left(2\right)$

Potential at the centre of the tunnel,
$V_f=\frac{V_s}{2}(3-\frac{r^2}{R^2})$

where, potential at the surface,
$V_s=\frac{\rho R^2}{3{ϵ}_0}$ and $r=\frac{R}{2}$
so,
$V_f=\frac{\rho R^2}{6{ϵ}_0}(3-\frac{R^2}{4R^2})$

$\therefore V_f=\frac{11\rho R^2}{24{ϵ}_0}....\left(3\right)$

Now from $\left(1\right),\left(2\right)$ and $\left(3\right)$, we get

$\frac{1}{2}m{v}^2-0=1[\frac{11\rho R^2}{24{ϵ}_0}-\frac{\rho R^2}{3{ϵ}_0}]$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{\rho R^2}{3{ϵ}_0}[\frac{11}{24}-\frac{1}{3}]$

$\Rightarrow \frac{1}{2}m{v}^2=\frac{\rho R^2}{8{ϵ}_0}$

Therefore the minimum velocity will be,

$v_{min}=v=\sqrt{\frac{\rho R^2}{4m{ϵ}_0}}$

Hence velocity should be slightly greater than $v$.

Hence option (a) is correct answer.