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Two point charges each of 5 muC but oppositein sign are placed 4cmfrom the mid pointon the axialline of dipole. |
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Answer» `2a = 4cm = 4xx10^(-2)m, E = ?` `r = 4 cm = 4xx10^(-2) m`. `E_("axial") = (2 PR)/(4 pi in_(0) (r^(2) - a^(2))^(2)) = (2 (q) (2a) r)/(4PI in_(0) (r^(2) - a^(2))^(2))` `= (9xx10^(9)xx2(5xx10^(-6))(4xx10^(-2)) xx4xx10^(-2))/([0.04^(2) - 0.02^(2)]^(2))` `E_("axial") = 10^(8) NC^(-1)` |
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