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Find the electric field intensity at 10cm away from a point charge of 100 esu.(a) 1 dyne/esu(b) 10 dyne/esu(c) 100 dyne/esu(d) 9*10^9 dyne/esuThe question was asked in examination.This interesting question is from Electric Field in section Charges and Fields of Physics – Class 12

Answer»

The CORRECT option is (a) 1 dyne/esu

To elaborate: We know that electric field intensity at r distance from a point charge q is DEFINED as \(\frac {q}{r^2}\) in the CGS SYSTEM. Here q=100 esu, r=10cm. Substituting the value we get E=1dyne/esu. We must remember that factor \(\frac {1}{4\pi\varepsilon}\) is CONSIDERED only in the SI system and its value is 9*10^9, but in the CGS system, we simply take the value of this constant as unity.



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