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A METAL SURFACE HAS A POSITIVE CHARGE OF 10-9 C HOW MANY ELECTRONS WOULD HAVE BEEN REMOVED FROM THE METAL SURFACE? A METAL SURFACE HAS A POSITIVE CHARGE OF 10-9 C HOW MANY ELECTRONS WOULD HAVE BEEN REMOVED FROM THE METAL SURFACE?
lol. i am also searchng for the answer Lekshmi :P
charge=q=10^-9 C , according to quantisation of charge we have, q = ne , n=no. of electrons, e=charge on 1 electron i.e. 1.6 * 10^-19 C, then n=q/e= 6.25 * 10^-29 here * stands for multiply. that`s all.
according to quantisation of charge, q =ne, q=charge ,n= no. of electrons, e= charge on an electron i.e. 1.6*10^-19, So n=q/e= 6.25*10^-29 * stands for multiply that`s all.
The elementary charge is 1.6x10^-19 . Thus to find the number of electrons removed from the surface divide the total charge by the elementary charge i.e. 10^-9/(1.6x10^-19) by using the formula q=ne therefore the final answer will be that 6.25x10^9 electrons have been removed.
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