Question

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/esu

The question was asked in examination.

This interesting question is from Electric Field in section Charges and Fields of Physics – Class 12

Answer 1

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.