Question

Four point charges $+8\mu C,-1\mu C,+8\mu C$ and $-1\mu C$ are fixed at the points $-\sqrt{27/2}m,-\sqrt{3/2}m,+\sqrt{3/2}m$ and $\sqrt{27/2}m$ respectively, on the y-axis. A particle of mass $6\times {10}^{-4}kg$ and charge +0.1$\mu C$ moves along the x-direction. Its speed at x = +$\infty$ is $V_o$. Find the least value of $V_o$ for which the particle will cross the origin. Find also the kinetic energy of the particle at the origin. Assume that space is gravity free. Given $\left(1/4\pi {\epsilon }_0\right)=9\times {10}^{9}N{m}^2{C}^{-2}$.

Answer 1

$q=10\mu C={10}^{-6}C{q}_0=0.1\mu C={10}^{-7}Cm=6\times {10}^{-4}kgQ=8\mu C=8\times 10-6C$

Let P be any point at a distance x from origin, then

$AP=CPBP=DP$

Electric potential at P will be

$V=\frac{2kQ}{BP}-\frac{2kq}{AP},\text{where}k=\frac{1}{4\pi {ϵ}_0}V=2k[\frac{Q}{{(\frac{27}{2}+x^2)}^{\frac{1}{2}}}-\frac{q}{{(\frac{3}{2}+x^2)}^{\frac{1}{2}}}]$

$V=2\times 9\times {10}^9\times {10}^{-6}[\frac{8}{{(\frac{27}{2}+x^2)}^{\frac{1}{2}}}-\frac{x}{{(\frac{3}{2}+x^2)}^{\frac{1}{2}}}]$

Electric field at P is

$E=-\frac{dV}{dx}=-1.8\times {10}^4[8\times (-\frac{1}{2})\times \frac{2x}{{(\frac{27}{2}+x^2)}^{\frac{3}{2}}}+\frac{1}{2}\frac{2x}{{(\frac{3}{2}+x^2)}^{\frac{3}{2}}}]E=1.8\times {10}^4[\frac{8x}{{(\frac{27}{2}+x^2)}^{\frac{3}{2}}}-\frac{x}{{(\frac{3}{2}+x^2)}^{\frac{3}{2}}}]$

$E=0$ at

$\frac{8x}{{(\frac{27}{2}+x^2)}^{\frac{3}{2}}}=\frac{x}{{(\frac{3}{2}+x^2)}^{\frac{3}{2}}}⟹\frac{4}{(\frac{27}{2}+x^2)}=\frac{1}{(\frac{3}{2}+x^2)}⟹\frac{4\times 3}{2}+4x^2=\frac{27}{2}+x^2⟹3x^2=\frac{27-2}{2}=\frac{15}{2}⟹x=\pm \sqrt{\frac{5}{2}}$

The least value of kinetic energy of the particle at infinity should be enough to take the particle upto $x=+\sqrt{\frac{5}{2}}$ beacause

at $x=+\sqrt{\frac{5}{2}},E=0$ Electrostatic force on $q_0$ is zero

for $x>\sqrt{\frac{5}{2}},E$ is repulsive (towards +ve x-axis)

for $x<\sqrt{\frac{5}{2}},E$ is attractive (towards -ve x-axis)

Now potential at $\sqrt{\frac{5}{2}}m$ is

$V=1.8\times {10}^4[\frac{8}{{(\frac{27}{2}+\frac{5}{2})}^{\frac{1}{2}}}-\frac{x}{{(\frac{3}{2}+\frac{5}{2})}^{\frac{1}{2}}}]V=1.8\times {10}^4\times \frac{3}{2}⟹V=2.7\times {10}^{4}V$

Applying energy conservation at $x=\infty$ and $x=\sqrt{\frac{5}{2}}m$

$\frac{1}{2}m{\nu }_0^2=q_{0}V⟹{\nu }_0=\sqrt{\frac{2q_{0}V}{m}}⟹{\nu }_0=\sqrt{\frac{2\times {10}^{-7}\times 2.7\times {10}^4}{6\times {10}^{-4}}}=3m/s$

Hence, minimum value of ${\nu }_0=3m/s$

Potential at origin is

$V_0=1.8\times {10}^4[\frac{8}{{\left(\frac{27}{2}\right)}^{\frac{1}{2}}}-\frac{x}{{\left(\frac{3}{2}\right)}^{\frac{1}{2}}}]⟹V-0=1.8\times {10}^4[\frac{8}{3\sqrt{\frac{3}{2}}}-\frac{1}{\sqrt{\frac{3}{2}}}]⟹V_0=1.8\times {10}^4\times \sqrt{\frac{2}{3}}\times \frac{5}{3}⟹V_0=\sqrt{6}\times {10}^4\approx 2.44\times {10}^4$

Let T be the kinetic energy of the particle at origin .

Applying energy conservation at x=0 and at $x=\infty$

$T+q_{0}V-0={\frac{m}{V}}_0^2⟹T=\frac{1}{2}m{V}_0^2-q_0{V}_0⟹T=\frac{1}{2}\times 6\times {10}^{-4}\times 3^2-2.44\times {10}^4\times {10}^{-7}⟹T=27\times {10}^{-4}-2.44\times {10}^{-4}⟹T=2.6\times {10}^{-4}J$