Question

Given point charges $+8\, \mu C ,-1 \, \mu C , -$ $1 \, \mu C$ and $+8\, \mu C$ are fixed at the points $-$ $\sqrt{\frac{27}{2}} m ,-\sqrt{\frac{3}{2}} m ,+\sqrt{\frac{3}{2}} m$ and $+\sqrt{\frac{27}{2}} m$ respectively on the $y$-axis, A particle of mass $6 \times 10^{-4} \, kg$ and charge $+0.1 \mu C$ moves along the $x$-axis. If Its speed at $x$ $=+\infty$ is $v_0$. Then find the minimum value of $v_0$ in $m s ^{-1}$ for which the particle will cross the origin. $\left(\frac{1}{4} \pi \varepsilon_0=9 x\right.$ $\left.10^9\, Nm ^2 C ^{-2}\right)$

Answer 1

In the figure $q =1 \mu C =10^{-6} C , q _0=+0.1 \mu C =$ $10^{-7} C$
and $m =6 \times 10^{-4} kg$ and $Q =8 \mu C =8 \times$
Let $P$ be any point at a distance $x$ from origin $O$. Then
$AP = CP =\sqrt{\frac{3}{2}+ x ^2}$
$BP = DP =\sqrt{\frac{27}{2}+ x ^2}$
Electric potential at point $P$ will be
$V =\frac{2 KQ }{ BP }-\frac{2 Kq }{ AP }$
where $K=\frac{1}{4 \pi \varepsilon_0}$
$=9 \times 10^9 Nm ^2 / C ^2$
Therefore $V =2 \times 9 \times 10^9 \frac{8 \times 10^{-6}}{\sqrt{\frac{27}{2}+ x ^2}}-\frac{10^{-6}}{\sqrt{\frac{3}{2}+ x ^2}}$
$V =1.8 \times 10^4 \frac{8}{\sqrt{\frac{27}{2}+ x ^2}}-\frac{10^{-6}}{\sqrt{\frac{3}{2}+ x ^2}} \ldots \text { (i) }$
Electric field at $P$ is
$E=-\frac{ dv }{ dx }=1.8 \times 10^4$
$8 \frac{-1}{2} \frac{27}{2}+x^{2^{-\frac{3}{2}}}--\frac{1}{2} \frac{3}{2}+x^{2^{-\frac{3}{2}}} 2 x$
$E=0$ on $x$-axis where $x=0$ or
$\frac{8}{\frac{27}{2}+ x ^{2^{\frac{3}{2}}}}=\frac{1}{\frac{3}{2}+ x ^{2^{\frac{3}{2}}}}$
$\Rightarrow \frac{4^{\frac{3}{2}}}{\frac{27}{2}+ x ^{2^{\frac{3}{2}}}}=\frac{1}{\frac{3}{2}+ x ^2 \frac{3}{2}}$
$\Rightarrow \frac{27}{2}+ x ^2=4 \frac{3}{2}+ x ^2$
This equation gives $x =\pm \sqrt{\frac{5}{2}} m$
The least value of kinetic energy of the particle at infinity should be enough to take the particle upto $x=+\sqrt{\frac{5}{2}} m$
because at $x =+\sqrt{\frac{5}{2}} m , E =0$
$\Rightarrow$ Electrostatic force on charge $q$ is zero or $F _{ e }=0$
For at $x>\sqrt{\frac{5}{2}} m , E$ is repulsive (towards positive $x$ - axis)
and for $x<\sqrt{\frac{5}{2}} m , E$ is attractive (towards negative $x$ - axis)
Now, from Eq. (i), potential at $x=\sqrt{\frac{5}{2}} m$
$V=1.8 \times 10^4 \frac{8}{\sqrt{\frac{27}{2}+\frac{5}{2}}}-\frac{1}{\sqrt{\frac{3}{2}+\frac{5}{2}}}$
$V=2.7 \times 10^4 \text { volt }$
Applying energy conservation at $x=\infty$ and $x=\sqrt{\frac{5}{2}} m$
$\frac{1}{2} mv ^2{ }_0= q _0 Vp \ldots \text { (ii) }$
$v _0=\sqrt{\frac{2 q_0 V}{m}}$
Substituting the values $v _0=$
$\sqrt{\frac{2 \times 10^{-7} \times 2.7 \times 10^4}{6 \times 10^{-4}}}$
$v_0=3 m / s$ Therefore Minimum value of $v_0$ is $3 m / s$