Question

How many electrons one has to remove from a neutral body to leave with a positive charge of 4.8×10^-8?

Answer 1

Given that,

Net positive charge, $q=4.8\times 10^{-8}\ C$

To find,

No of electrons removed.

Solution,

Let n electrons are removed. Using quantization of electric charge to find n as  follows :

q=ne

Where, e is charge on an electron

$n=\dfrac{q}{e}\\\\n=\dfrac{4.8\times 10^{-8}}{1.6\times 10^{-19}}\\\\n=3\times 10^{11}$

So, $3\times 10^{11}$ electrons has to remove from a neutral body.