CBSE Class 12-science Answered

Derivation of electric field on axial and equatorial of electric dipole

Asked by sonunahid64 | 08 Apr, 2021, 07:27: AM

Expert Answer

Electric field due to dipole in axial direction is determined using using above figure.

Electric field at a point P(x,0) is given as

begin mathsize 14px style E space equals space E subscript plus minus space E subscript minus space equals space K space q over left parenthesis x minus d right parenthesis squared minus space K space q over left parenthesis x plus d right parenthesis squared equals space K q space fraction numerator 2 x d over denominator left parenthesis space x squared minus d squared space right parenthesis end fraction equals space K space fraction numerator left parenthesis q cross times 2 d right parenthesis space x over denominator left parenthesis space x squared minus d squared space right parenthesis end fraction space equals space K fraction numerator p space x over denominator left parenthesis space x squared minus d squared space right parenthesis end fraction end style

Where K = 1 / (4πε_o ) is Coulomb's constant and p = ( q × 2d ) is dipole moment

Electric field of dipole in equtorial direction is determined using above figure

Electric field E₊ due to positive charge is given as

begin mathsize 14px style E subscript plus space equals space K space fraction numerator q over denominator open parentheses y squared plus d squared over 4 close parentheses end fraction end style

Electric field E_- due to negative charge is given as

begin mathsize 14px style E subscript minus space equals space K space fraction numerator q over denominator open parentheses y squared plus d squared over 4 close parentheses end fraction end style

Since magnitudue of above fields are same , resultant field E is given as

begin mathsize 14px style E space equals space 2 space open vertical bar E subscript plus close vertical bar space cos theta space equals space 2 space K space fraction numerator q over denominator open parentheses y squared plus d squared over 4 close parentheses end fraction cross times fraction numerator d divided by 2 over denominator square root of y squared plus d squared over 4 end root end fraction space equals space K space fraction numerator p over denominator square root of 4 y squared plus d squared end root end fraction end style

Answered by Thiyagarajan K | 08 Apr, 2021, 12:00: PM

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