CBSE Class 12-science Answered
Plzz explain
![question image](https://images.topperlearning.com/topper/new-ate/topr_893561764621337314.jpeg)
Asked by dharaelect | 27 Mar, 2019, 13:58: PM
Two charges of +10 μC and - 10 μC are placed at a distance of 5 mm.
Dipole moment is given as,
p = qd
p = 10 x 10-6 x 5 x 10-3
p = 5 x 10-8 Cm
a) Point P is on axis at a distance of 15 cm from a distance O.
Thus, the electric field at point P is,
![E subscript P equals fraction numerator 2 p over denominator 4 pi epsilon subscript 0 r cubed end fraction space equals space fraction numerator 9 space x space 10 to the power of 9 space cross times space 2 space cross times space 5 space cross times space 10 to the power of negative 8 end exponent over denominator left parenthesis 15 space cross times space 10 to the power of negative 2 end exponent right parenthesis cubed end fraction space equals space 2.666 space cross times space 10 to the power of 5 space N divided by C space almost equal to space 2.67 space cross times space 10 to the power of 5 space N divided by C](https://images.topperlearning.com/topper/tinymce/cache/19911c719645f5f8d91771a5a2813f97.png)
b) The point Q is perpendicular to the axis of the dipole which is 15 cm apart from point O.
Thus, the electric field due to a dipole at Q is,
![E space equals space fraction numerator p over denominator 4 pi epsilon subscript 0 r cubed end fraction space
T h u s comma space
E subscript Q space equals space E subscript P over 2 space equals space fraction numerator 2.67 space cross times space 10 to the power of 5 over denominator 2 end fraction space equals space 1.33 space cross times space 10 to the power of 5 space N divided by C](https://images.topperlearning.com/topper/tinymce/cache/d2bd72ede3b586236eb89aa5878cfcab.png)
Answered by Shiwani Sawant | 27 Mar, 2019, 15:44: PM
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