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CBSE Class 12-science Answered

Plzz explain
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Asked by dharaelect | 27 Mar, 2019, 01:58: PM
answered-by-expert Expert Answer
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
 
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
Answered by Shiwani Sawant | 27 Mar, 2019, 03:44: PM

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