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NEET Class neet Answered

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Asked by rekhapremkumar25 | 27 Aug, 2021, 07:06: AM
Expert Answer
Distance dn of nth charge from origin is given as
 
dn = 2(n-1) m
 
Electric field E due to infinite number of charges is given as
 
begin mathsize 14px style E space equals space sum from n equals 1 to infinity of K space cross times q over 2 to the power of left parenthesis 2 n minus 2 right parenthesis end exponent space N divided by C end style
where K = 1/(4πεo ) is Coulomb's constant
 
Hence E = K q [ 1 + (1/4) + (1/16) +(1/64) ....................... to ∞ ]  N/C = K q Sn N/C
 
Bracketted term in above expression is geometric series with first term is 1 and common multipier is (1/4) .
 
Sn is sum of geometric series
 
Sn =  1 / [ 1 - (1/4) ] = 4/3   m-2
 
Hence electric field E = (4/3) K q
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