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CBSE Class 12-science Answered

a copper wire ring of radius 15cm is given a charge 100uc.find the electric potential and the intensities at the center of the ring and the point on the axis at a distance 10cm from its center  
Asked by bhushankr113 | 13 Jun, 2019, 10:20: PM
answered-by-expert Expert Answer
At point O :
electric potential = begin mathsize 14px style fraction numerator q over denominator 4 πε subscript straight o straight r end fraction space equals space space 9 cross times 10 to the power of 9 space fraction numerator 100 cross times 10 to the power of negative 6 end exponent over denominator 15 cross times 10 to the power of negative 2 end exponent end fraction space equals space 6 cross times 10 to the power of 6 space V end style
electric field = begin mathsize 14px style fraction numerator q over denominator 4 πε subscript straight o straight r squared end fraction space equals space space 9 cross times 10 to the power of 9 space fraction numerator 100 cross times 10 to the power of negative 6 end exponent over denominator 225 cross times 10 to the power of negative 4 end exponent end fraction space equals space 4 cross times 10 to the power of 7 space N divided by C end style
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At point P, i.e. 15 cm from centre of ring in the axis
 
electric potential = begin mathsize 14px style fraction numerator q over denominator 4 πε subscript straight o space A P end fraction space equals space space 9 cross times 10 to the power of 9 space fraction numerator 100 cross times 10 to the power of negative 6 end exponent over denominator square root of 325 cross times 10 to the power of negative 2 end exponent end fraction space equals space 5 cross times 10 to the power of 6 space V end style
field vectors are shown in figure due to charge elements located opposite points in the ring.
If we resolve the vectors, vertical components will be nullified. Horizontal components are getting added up.
angle θ = tan-1(10/15)≈56
electric field = begin mathsize 14px style fraction numerator q over denominator 4 πε subscript straight o space A P squared end fraction space cos theta space equals space space 9 cross times 10 to the power of 9 space fraction numerator 100 cross times 10 to the power of negative 6 end exponent over denominator 325 cross times 10 to the power of negative 4 end exponent end fraction cross times cos left parenthesis 56 right parenthesis space almost equal to space 1.54 cross times 10 to the power of 7 space N divided by C end style
 
 
Answered by Thiyagarajan K | 14 Jun, 2019, 12:28: PM

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