A) \[x=2\] and \[x=9\]
B) \[x=1\] and \[x=5\]
C) \[x=4\] and \[x=12\]
D) \[x=-2\] and \[x=2\]
Correct Answer: C
Solution :
Potential will be zero at two points At internal point (M) :\[\frac{1}{4\pi {{\varepsilon }_{0}}}\times \left[ \frac{2\times {{10}^{-6}}}{(6-l)}+\frac{(-1\times {{10}^{-6}})}{l} \right]=0\] Þ l = 2 So distance of M from origin; x = 6 ? 2 = 4 At exterior point (N) :\[\frac{1}{4\pi {{\varepsilon }_{0}}}\times \left[ \frac{2\times {{10}^{-6}}}{(6-l')}+\frac{(-1\times {{10}^{-6}})}{{{l}^{'}}} \right]=0\] Þ \[l'=6\] So distance of N from origin, x = 6 + 6 = 12You need to login to perform this action.
You will be redirected in
3 sec