In fig three point charges q,-2q are q are placed along the x axis. Show that the electric field at p along the y axis is E =(1/4×3.14×e knot)×(3qa^2/y^4),when y>>a.integral to the power of straight R to the power of straight y end exponent over blank

Asked by sharmashiva3667.12sdatl | 8th Jun, 2020, 08:32: PM

Expert Answer:

r space equals space square root of y squared plus a squared end root L e t space E subscript 1 space b e space t h e space e l e c t r i c space f i e l d space a t space P space d u e space t o space q space E subscript 1 space equals space open parentheses fraction numerator 1 over denominator 4 pi epsilon subscript 0 end fraction space close parentheses q over r squared space L e t space E subscript 2 space b e space t h e space e l e c t r i c space f i e l d space a t space P space d u e space t o space minus space 2 q space T h u s comma space E subscript 2 space equals open parentheses space fraction numerator 1 over denominator 4 pi epsilon subscript 0 end fraction close parentheses fraction numerator 2 q over denominator y squared end fraction space T h u s comma space n e t space e l e c t r i c space f i e l d space a t space P space E thin space equals space 2 space E subscript 1 cos theta space plus thin space E subscript 2 equals space 2 E subscript 1 space cos theta space plus thin space E subscript 2 space equals space 2 open parentheses space fraction numerator 1 over denominator 4 pi epsilon subscript 0 end fraction close parentheses open parentheses q over y squared space close parentheses open parentheses y over r close parentheses space plus open parentheses space fraction numerator 1 over denominator 4 pi epsilon subscript 0 end fraction close parentheses fraction numerator 2 q over denominator y squared end fraction space space equals space open parentheses space fraction numerator 2 q over denominator 4 pi epsilon subscript 0 end fraction close parentheses space open parentheses y over open parentheses a squared plus thin space y squared close parentheses to the power of begin display style 3 over 2 end style end exponent space plus thin space 1 over y squared close parentheses space A p p l y i n g space b i o n o m i a l space e x p a n s i o n space f o r space a squared over y squared space less than less than 1 E space space equals space space open parentheses space fraction numerator 2 q over denominator 4 pi epsilon subscript 0 end fraction close parentheses open parentheses fraction numerator negative 3 a squared over denominator 2 y to the power of 4 end fraction close parentheses space equals space minus open parentheses space fraction numerator 3 a squared q over denominator 4 pi epsilon subscript 0 y to the power of 4 end fraction close parentheses space

Answered by Shiwani Sawant | 9th Jun, 2020, 09:10: PM

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