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CBSE Class 12-science Answered

derive an expression for electric field intensity at a point ont he eqiulatory line of the dipole.also determine its direction.
Asked by Balan | 26 Mar, 2018, 01:04: PM
answered-by-expert Expert Answer
As shown in the figure, E+ is the electric field at a distance r from centre of dipole due to charge +q. Similarly E- is the field due to charge -q.
It can be seen from figure that vertical componenet will cancel each other and the resultant field 2Ecosθ is in horizontal direction parallel to horizontal.
where E = |E+| = |E-| and angle θ is shown in the figure.
hence net field is given by
begin mathsize 12px style E space equals fraction numerator 2 q over denominator 4 pi epsilon subscript 0 open parentheses r squared plus d squared close parentheses end fraction space cos space theta equals space fraction numerator 2 q over denominator 4 pi epsilon subscript 0 r squared end fraction open parentheses 1 plus d squared over r squared close parentheses to the power of negative 1 end exponent d over r space equals fraction numerator 2 q d over denominator 4 pi epsilon subscript 0 r cubed end fraction open parentheses 1 minus d squared over r squared close parentheses sin c e space d over r less than less than 1 comma space n e t space F i e l d space space E space equals space fraction numerator 2 q d over denominator 4 pi epsilon subscript 0 r cubed end fraction equals space fraction numerator p over denominator 4 pi epsilon subscript 0 r cubed end fraction end style
where p = 2qd, is dipole moment . Direction of the resultant field is shown in diagram
Answered by Thiyagarajan K | 27 Mar, 2018, 07:24: AM

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