A) 5 cm
B) \[\frac{25}{9}\]cm
C) 10 cm
D) \[\frac{4}{13}\] cm
Correct Answer: A
Solution :
Suppose neutral point N lies at a distance x from dipole of moment p or at a distance x2 from dipole of 64 p. At N |E. F. due to dipole |= |E. F. due to dipole | Þ \[\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2p}{{{x}^{3}}}=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{2(64p)}{{{(25-x)}^{3}}}\] Þ \[\frac{1}{{{x}^{3}}}=\frac{64}{{{(25-x)}^{3}}}\] Þ x = 5 cm.You need to login to perform this action.
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