two point charge 2microcoulomb and 6 microcoulomb repel each other with a force of 12N if each is given a additional charge of -4microcoulomb ,what will be the new force?

Asked by Subhashree Lenka | 23rd May, 2015, 10:22: AM

Expert Answer:

begin mathsize 14px style We space know space that space the space force space between space two space charges space is space given space by space coloumbs space law colon straight F equals fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator straight q subscript 1 straight q subscript 2 over denominator straight r squared end fraction Given space that space straight F equals space 12 space straight N comma space straight q subscript 1 space equals 2 cross times 10 to the power of negative 6 end exponent space straight C comma space straight q subscript 2 space equals space 6 cross times 10 to the power of negative 6 end exponent space straight C straight F equals fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator open parentheses 2 cross times 10 to the power of negative 6 end exponent close parentheses cross times open parentheses space 6 cross times 10 to the power of negative 6 end exponent close parentheses over denominator straight r squared end fraction 12 space equals 9 cross times 10 to the power of 9 cross times fraction numerator open parentheses 2 cross times 10 to the power of negative 6 end exponent close parentheses cross times open parentheses space 6 cross times 10 to the power of negative 6 end exponent close parentheses over denominator straight r squared end fraction straight r squared space equals 9 cross times 10 to the power of negative 3 end exponent When space an space additional space charge space is space given space the space force space will space be colon straight F equals fraction numerator 1 over denominator 4 πε subscript 0 end fraction fraction numerator open parentheses 2 minus 4 cross times 10 to the power of negative 6 end exponent close parentheses cross times open parentheses space 6 minus 4 cross times 10 to the power of negative 6 end exponent close parentheses over denominator straight r squared end fraction straight F equals 9 cross times 10 to the power of 9 cross times fraction numerator open parentheses negative 2 cross times 10 to the power of negative 6 end exponent close parentheses cross times open parentheses space 2 cross times 10 to the power of negative 6 end exponent close parentheses over denominator 9 cross times 10 to the power of negative 3 end exponent end fraction straight F equals negative space 4 space straight N Therefore space the space new space force space equals space minus 4 space straight N end style

Answered by Jyothi Nair | 24th May, 2015, 03:21: PM

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