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# An electron revolves around the nucleus of hydrogen atom in a circle of radius 5×10−11 m. The intensity of electric field at a point in the orbit of the electron is:9.216×10−8N/C5.76×1011N/C04N/C

A
5.76×1011N/C
B
4N/C
C
9.216×108N/C
D
0
Solution
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#### We know hydrogen atom consist of one proton and one electronR=5×10−11mWe know attractive force =kq1(q2)r2=9×109×(1⋅6×10−19)2(5×10−11)2We know F=Eq∴E=Fq∴E=9×109×(1⋅6×10−19)5×10−11)2=5⋅76×1011N/C.

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