CBSE Class 12-science Answered
calculate the electric field due to the electric dipole of length 2 cm having charge 4 microcoulomb at an angle of minus 60 degree for the direction of electric dipole moment 1 meter for the center to the dipole
Asked by surajsinghrajput4691 | 16 May, 2022, 00:20: AM
![](https://images.topperlearning.com/topper/tinymce/imagemanager/files/09083bca184454fd3994556529d841b16281c8339d49d7.09448151f6.png)
Let the dipole is at origin of cartesian coordinate system as shown in figure.
electric field at P is given as
![begin mathsize 14px style E space equals space fraction numerator q over denominator 4 pi epsilon subscript o end fraction open parentheses fraction numerator 1 over denominator r subscript plus superscript 2 end fraction space minus space fraction numerator 1 over denominator r subscript minus superscript 2 end fraction close parentheses end style](https://images.topperlearning.com/topper/tinymce/cache/ee85fadbc7370ebc220616546aaff2f0.png)
where q is magnitude of individual charge of dipole , r+ is ditsance between positive charge and point P ,
r- is ditsance between negative charge and point Pand εo is permittivity of free space .
distance r+ is determined as follows using cosine rule
![begin mathsize 14px style r subscript plus superscript 2 space equals space r to the power of 2 to the power of space end exponent plus d squared space minus space 2 space r space d space cos 120 space equals space r to the power of 2 to the power of space end exponent plus d squared space plus space space r space d space
end style](https://images.topperlearning.com/topper/tinymce/cache/5457a1c7935ea5a30e08b423a2af5f38.png)
where r is distance from origin to point P and 2d is length of dipole
![begin mathsize 14px style r subscript plus superscript 2 space equals space r to the power of 2 to the power of space end exponent open parentheses 1 space plus stretchy left parenthesis d over r stretchy right parenthesis squared space plus space space d over r close parentheses space almost equal to r to the power of 2 to the power of space end exponent stretchy left parenthesis 1 space space plus space space d over r stretchy right parenthesis space
end style](https://images.topperlearning.com/topper/tinymce/cache/971a3c39f9854a168ca8c2afed590d37.png)
![begin mathsize 14px style fraction numerator 1 over denominator r subscript plus superscript 2 end fraction space equals space 1 over r squared space open parentheses 1 plus d over r close parentheses to the power of negative 1 end exponent space equals space 1 over r squared space open parentheses 1 minus d over r close parentheses end style](https://images.topperlearning.com/topper/tinymce/cache/50834d1dcbb79e0ba54e873f9bce3be2.png)
similarly we get
![begin mathsize 14px style fraction numerator 1 over denominator r subscript minus superscript 2 end fraction space equals space 1 over r squared space open parentheses 1 plus d over r close parentheses end style](https://images.topperlearning.com/topper/tinymce/cache/fbc1b2e7acd3ca7da73fa4361558d627.png)
![begin mathsize 14px style E space equals space fraction numerator negative 2 q over denominator 4 pi epsilon subscript o space r squared end fraction open parentheses d over r close parentheses space space equals space fraction numerator negative 2 q d over denominator 4 pi epsilon subscript o r cubed end fraction end style](https://images.topperlearning.com/topper/tinymce/cache/9687a10e204e9bb828150eae5e42fc7e.png)
let us substitute values
1/(4πεo ) = 9 × 109 N m2 C-2 , q = 4 × 10-6 C and d =1 cm
E = - ( 9 × 109 × 2 × 4 × 10-6 × 0.01 ) / 13 = -720 N
Answered by Thiyagarajan K | 16 May, 2022, 09:57: AM
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