c
(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 \((N)\) :\(\frac{1}{{4\pi {\varepsilon _0}}}\) \(×\) \(\left[ {\frac{{2 \times {{10}^{ - 6}}}}{{6 - l'}} + \frac{{ - 1 \times {{10}^{ - 6}}}}{{l'}}} \right] =0 \)
\(l'=6\)
So distance of \(N\) from origin, \(x = 6 + 6 = 12\)
