IIT JEE , AIEEE Forum - Physics , Electricity

akumar_ak

the nucleus of a hydrogen atom is a sphere of radius 10^-15 m.The electric field at the surface of the nucleus is:
a)14.4x10⁴ Vm^-1
b)14.4x10^-1Vm-1
c)14.4x10¹⁹Vm^-1
d)14.4x10²⁰ Vm^-1according to me the answer is (a)
and u too calculate,i'm sure you'd agree with me
but according to the book,its given (d)

verify if u get that

biki

Why will u get that answer fried..???

Electric field = 9x10⁹ x q/r²

Here the entire charge is due to a single proton dear...as the field is to be calculated "on the surface of the nucleus"

whose charge = that of electron = q = 1.6x10^-19 C

and r = 10^-15 m

Put the values....

E. field = 9x10⁹x 1.6x10^-19 / (10^-15)²

= 14.4x10²⁰ N/C

= 14.4x10²⁰V m^-1....

friend ,,, i think you had some calculation mistake....or else its easy.!!!

akumar_ak

ohhhhhhh see,i didn't square the distance

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