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CBSE Class 12-science Answered

A body at rest of mass 4 pound and charge 3 esu placed inside uniform electric field intensity 4N/C. Find the velocity of the charge body after 2 sec.
Asked by pradeepjsme | 26 Jun, 2018, 12:05: AM
answered-by-expert Expert Answer
Force F = q×E begin mathsize 12px style equals space 3 space e s u space cross times fraction numerator 1 space C over denominator 3 cross times 10 to the power of 9 e s u end fraction cross times 4 space N over C space equals space 4 cross times 10 to the power of negative 9 end exponent space N end style
acceleration a = Force / mass begin mathsize 12px style equals space fraction numerator 4 cross times 10 to the power of negative 9 end exponent space N over denominator 4 space P o u n d s end fraction cross times fraction numerator 1 space P o u n d over denominator 0.454 space k g end fraction space equals space 2.2 cross times 10 to the power of negative 9 end exponent space m divided by s squared end style
velocity after 2 sec starting from rest = a×t = 2.2×10-9×2 = 4.4 × 10-9 m/s
Answered by Thiyagarajan K | 26 Jun, 2018, 10:41: AM

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