1.

Two point charges each of `5 muC` but opposite in sign are placed 4cm from the mid point on the axial line of dipole.

Answer» Correct Answer - `10^(8) NC^(-1)`
Here, `q = 5 muC = 5xx10^(-6)C`
`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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