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CBSE Class 12-science Answered

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Asked by fishtailfever | 18 Mar, 2020, 12:57: PM
answered-by-expert Expert Answer
 
Force F1 due to charge q1 at point B is given as, 
F subscript 1 equals k fraction numerator q subscript 1 q subscript 2 over denominator r squared end fraction space... left parenthesis r e p u l s i v e space f o r c e right parenthesis space equals fraction numerator 9 cross times 10 to the power of 9 cross times 1.5 cross times 10 to the power of negative 3 end exponent cross times 0.2 cross times 10 to the power of negative 3 end exponent over denominator 1.2 squared end fraction F subscript 1 equals 1.875 cross times 10 cubed space N thin space F subscript 2 space equals space minus k fraction numerator q subscript 2 space q subscript 3 over denominator r squared end fraction space... space left parenthesis a t t r a c t i v e space f o r c e right parenthesis equals space fraction numerator 9 cross times 10 to the power of 9 cross times 0.5 cross times 10 to the power of negative 3 end exponent cross times 0.2 cross times 10 to the power of negative 3 end exponent over denominator 0.6 squared end fraction equals space minus 2.5 space cross times space 10 cubed space N T h u s comma space F space equals space square root of F subscript 1 superscript 2 plus F subscript 2 superscript 2 2 F subscript 1 F subscript 2 space cos theta end root theta equals 90 cos 90 degree space equals 0 space F equals space square root of left parenthesis 1.875 right parenthesis squared plus left parenthesis 2.5 right parenthesis squared space x open parentheses 10 cubed close parentheses squared end root space equals space square root of 9.765625 end root space cross times space 10 cubed space equals space 3.125 space cross times space 10 cubed N  
 
Thus, the magnitude of force isspace 3.125 space cross times space 10 cubed N
Answered by Shiwani Sawant | 18 Mar, 2020, 02:43: PM

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