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CBSE Class 12-science Answered

A body at rest of mass 2mg and charge 2C placed inside uniform electric field intensity 2 dyne/ esu. If the distance travel by the charge body is 5m . Find the time taken by the charge body in this process.
Asked by Naira | 24 Jun, 2018, 04:23: PM
answered-by-expert Expert Answer
Electric field intensity begin mathsize 12px style equals space 2 space fraction numerator d y n e over denominator e s u end fraction cross times fraction numerator 10 to the power of negative 5 end exponent N over denominator d y n e end fraction cross times fraction numerator 3 cross times 10 to the power of 9 e s u over denominator C end fraction space equals 6 cross times 10 to the power of 4 space bevelled N over C end style
force experienced by 2 C charge = 2×6×104 N/C= 12×104 N/C
 
acceleration of mass 2 mg = begin mathsize 12px style fraction numerator F o r c e over denominator m a s s end fraction space equals space fraction numerator 12 cross times 10 to the power of 4 over denominator 2 cross times 10 to the power of negative 3 end exponent end fraction space equals space 6 cross times 10 to the power of 7 space m divided by s squared end style
Time t to travel a distance 5 m starting from rest is obtained by using the formula "S = ut+(1/2)at2 ", with initial velocity u = 0 ;
 
hence we have,     begin mathsize 12px style 5 space equals space 1 half cross times 6 cross times 10 to the power of 7 cross times t squared space.................. left parenthesis 1 right parenthesis end style
solving for t using eqn.(1), we get t = 0.41 milli second
Answered by Thiyagarajan K | 24 Jun, 2018, 11:14: PM

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