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. $(\frac{1}{4}\pi {\epsilon }_0=9x$ $10^{9}N{m}^2{C}^{-2})$

Answer 1

Answer: 3

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
$V=\frac{2KQ}{BP}-\frac{2Kq}{AP}$
where $K=\frac{1}{4\pi {\epsilon }_0}$
$=9\times 10^{9}N{m}^2/C^2$
Therefore $V=2\times 9\times 10^9\frac{8\times 10^{-6}}{\sqrt{\frac{27}{2}+x^2}}-\frac{10^{-6}}{\sqrt{\frac{3}{2}+x^2}}$
$V=1.8\times 10^4\frac{8}{\sqrt{\frac{27}{2}+x^2}}-\frac{10^{-6}}{\sqrt{\frac{3}{2}+x^2}}\dots \text{ (i) }$
Electric field at $P$ is
$E=-\frac{dv}{dx}=1.8\times 10^4$
$8\frac{-1}{2}\frac{27}{2}+x^{2^{-\frac{3}{2}}}--\frac{1}{2}\frac{3}{2}+x^{2^{-\frac{3}{2}}}2x$
$E=0$ 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 $x=\sqrt{\frac{5}{2}}m$
$\frac{1}{2}m{v}^2{}_0=q_{0}Vp\dots \text{ (ii) }$
$v_0=\sqrt{\frac{2q_{0}V}{m}}$
Substituting the values $v_0=$
$\sqrt{\frac{2\times 10^{-7}\times 2.7\times 10^4}{6\times 10^{-4}}}$
$v_0=3m/s$ Therefore Minimum value of $v_0$ is $3m/s$