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Two charges `+20 muC and -20 muC` are held `1 cm` apart. Calculate the electric field at a point on the equatorial line at a distance of `50 cm` from the center of the dipole. |
Answer» Here, ` q = +- mu C = +- 20xx10^(-6) C`, `2a = 1 cm = 10^(-2)m, r = 50 cm = (1)/(2) m` As `2 a lt lt r`, therefore, intensity on equatorial line of short dipole is ` E = (1)/(4pi in_(0)) (p)/(r^(3)) = (qxx2a)/(4pi in_(0) r^(3))` `= (9xx10^(9)xx20xx10^(-6)xx10^(-2))/((1//2)^(3))` `E = 1*44xx10^(4) N//C` |
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