Point charge ${q_1} = 2\,\mu C$ and ${q_2} = - 1\,\mu C$ are kept at points $x = 0$ and $x = 6$ respectively. Electrical potential will be zero at points
Diffcult
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(c) Potential will be zero at two points
At internal point $(M)$ :$\frac{1}{{4\pi {\varepsilon _0}}} \times \left[ {\frac{{2 \times {{10}^{ - 6}}}}{{(6 - l)}} + \frac{{( - 1 \times {{10}^{ - 6}})}}{l}} \right] = 0$
$==>$ $l = 2$
So distance of $M$ from origin; $x = 6 -2 = 4$
At exterior point $(N)$ :$\frac{1}{{4\pi {\varepsilon _0}}}$ $×$ $\left[ {\frac{{2 \times {{10}^{ - 6}}}}{{6 - l'}} + \frac{{ - 1 \times {{10}^{ - 6}}}}{{l'}}} \right] =0 $
At internal point $(M)$ :$\frac{1}{{4\pi {\varepsilon _0}}} \times \left[ {\frac{{2 \times {{10}^{ - 6}}}}{{(6 - l)}} + \frac{{( - 1 \times {{10}^{ - 6}})}}{l}} \right] = 0$
$==>$ $l = 2$
So distance of $M$ from origin; $x = 6 -2 = 4$
At exterior point $(N)$ :$\frac{1}{{4\pi {\varepsilon _0}}}$ $×$ $\left[ {\frac{{2 \times {{10}^{ - 6}}}}{{6 - l'}} + \frac{{ - 1 \times {{10}^{ - 6}}}}{{l'}}} \right] =0 $
$l'=6$
So distance of $N$ from origin, $x = 6 + 6 = 12$
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