A) \[\frac{k.p}{{{x}^{3}}}\]
B) \[\frac{2kp}{{{x}^{3}}}\]
C) Zero
D) \[\frac{\sqrt{2}kp}{{{x}^{3}}}\]
Correct Answer: C
Solution :
Point P lies at equatorial positions of dipoles 1and 2 and axial position of dipole 3. Hence field at P due to dipole 1, \[{{E}_{1}}=\frac{k.p}{{{x}^{3}}}\,\,(towards\,\,left)\] due to dipole 2, \[{{E}_{2}}=\frac{k.p}{{{x}^{2}}}\,\,(towards\,\,left)\] due to dipole 3, \[{{E}_{3}}=\frac{k.(2p)}{{{x}^{3}}}\,\,(towards\,\,right)\] So net field at P will be zero.You need to login to perform this action.
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