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CBSE Class 12-science Answered

Two point charges 9 x 10 -8. and -9x10 -8 C are 0.5 m appart in air. Find the resultant electric intensity at a point located 0.5 m from either charge.
Asked by rutujasarangmehta | 04 Mar, 2016, 09:45: AM
answered-by-expert Expert Answer
Given:
q = ± 9 × 10-8 C
2 a = 0.5 m
r = 0.5 m
Electric field intensity on equitorial of dipole
begin mathsize 14px style straight E equals fraction numerator 1 over denominator 4 πε subscript 0 end fraction straight p over straight r cubed equals fraction numerator straight q space cross times space 2 straight a over denominator 4 πε subscript 0 straight r cubed end fraction space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 9 cross times 10 to the power of 9 cross times 9 cross times 10 to the power of negative 8 end exponent cross times 0.5 over denominator left parenthesis 0.5 right parenthesis cubed end fraction space space space space space space space space space space space space space space space space space space space space space space equals fraction numerator 81 cross times 10 to the power of 9 cross times 10 to the power of negative 8 end exponent over denominator left parenthesis 0.5 right parenthesis squared end fraction equals fraction numerator 81 cross times 10 to the power of 1 over denominator 0.25 end fraction equals fraction numerator 810 over denominator 0.25 end fraction equals 32.40 cross times 10 squared space straight N divided by straight C end style
Answered by Faiza Lambe | 07 Mar, 2016, 12:18: PM

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