Given point charges + 8 μC , - 1 μC ,   - 1 μC and + 8 μC   are fixed at the…

1. 1
2. 2
3. 3
4. 4

Solution

 
In the figure q = 1μC = 10^-6C, q₀ = + 0. 1μC = 10^-7C

and         m = 6x 10^-4 kg  and Q = 8μC = 8 x 10^-6 C

Let P be any point at a distance x from origin O. Then
 

$\text{AP}=\text{CP}=\sqrt{\frac{3}{2}+{\text{x}}^2}$  
 
$\text{BP}=\text{DP}=\sqrt{\frac{27}{2}+{\text{x}}^2}$

Electric potential at point P will be
 
$\text{V}=\frac{2\text{KQ}}{\text{BP}}-\frac{2\text{Kq}}{\text{AP}}$
 
 where $K=\frac{1}{4\pi {\epsilon }_0}$

= 9 x 10⁹ Nm²/C²
 
Therefore V = 2 x 9 x 10⁹ $\left[\frac{8\times 10^{-6}}{\sqrt{\frac{27}{2}+{\text{x}}^2}}-\frac{10^{-6}}{\sqrt{\frac{3}{2}+{\text{x}}^2}}\right]$
 V = 1. 8 x 10⁴ $\left[\frac{8}{\sqrt{\frac{27}{2}+{\text{x}}^2}}-\frac{{\text{10}}^{-6}}{\sqrt{\frac{3}{2}+{\text{x}}^2}}\right]$... (i)
 
 
Electric field at P is
$E=-\frac{\mathrm{d}\mathrm{v}}{\mathrm{d}\mathrm{x}}=1.8\times {10}^4\left[\left(8\right)\left(\frac{-1}{2}\right){\left(\frac{27}{2}+x^2\right)}^{-\frac{3}{2}}-\left(-\frac{1}{2}\right){\left(\frac{3}{2}+x^2\right)}^{-\frac{3}{2}}\right]\left(2x\right)$
E = 0 on x -axis where x = 0 or
 
$\frac{8}{{\left(\frac{27}{2}+{\text{x}}^2\right)}^{\frac{3}{2}}}=\frac{1}{{\left(\frac{3}{2}+{\text{x}}^2\right)}^{\frac{3}{2}}}$
 
$\Rightarrow \frac{{\left(4\right)}^{\frac{3}{2}}}{{\left(\frac{27}{2}+{\text{x}}^2\right)}^{\frac{3}{2}}}=\frac{1}{{\left(\frac{3}{2}+{\text{x}}^2\right)}^{\frac{3}{2}}}$
 

$\Rightarrow \left(\frac{27}{2}+{\text{x}}^2\right)=4\left(\frac{3}{2}+{\text{x}}^2\right)$
 
This equation gives $\text{x}=\pm \sqrt{\frac{5}{2}}\text{m}$

The least value of kinetic energy of the particle at infinity should be enough to take the particle upto x = +$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$\sqrt{\genfrac{}{}{0.1ex}{}{5}{2}}$m
 
because at x = +$\sqrt{\genfrac{}{}{0.1ex}{}{5}{2}}$m, E = 0.
 
=>  Electrostatic force on charge q is zero or F_e = 0.
 
For at x >$\sqrt{\genfrac{}{}{0.1ex}{}{5}{2}}$m, E is repulsive (towards positive x - axis)
 
and for x <$\sqrt{\genfrac{}{}{0.1ex}{}{5}{2}}$m, E is attractive (towards negative x - axis)
 
Now, from Eq. (i), potential at x =  $\sqrt{\genfrac{}{}{0.1ex}{}{5}{2}}$m

V = 1. 8x 10⁴ $\left[\frac{8}{\sqrt{\frac{27}{2}+\frac{5}{2}}}-\frac{1}{\sqrt{\frac{3}{2}+\frac{5}{2}}}\right]$
 
V = 2. 7 x 10⁴ volt

Applying energy conservation at x = ∞ and x =$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$\sqrt{\genfrac{}{}{0.1ex}{}{5}{2}}$m

$\genfrac{}{}{0.1ex}{}{1}{2}$  mv²₀ = q₀Vp ... (ii)
 
v₀ = $\sqrt{\genfrac{}{}{0.1ex}{}{2q_{0}V}{m}}$
Substituting the values v₀ = $\sqrt{\genfrac{}{}{0.1ex}{}{2\times 10^{-7}\times 2.7\times 10^4}{6\times 10^{-4}}}$
                                                                       
v₀ = 3 m/s Therefore Minimum value of v₀ is 3 m/s ,

Asked in: JEE Mains - Electrostatics - Test 2

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