Question

50 billion million deficit of electrons will have a charge of

Answer 1

First we confirm how many electrons were removed:

$N=\text{50 billion million electrons} \\N=(50)(10^9)(10^6)\text{ electrons} \\N=5\times10^{16}\text{ electrons}$

Since we are removing electrons (because this is a deficit), the total charge will be positive:

$Q=5\times10^{16}\,\cancel {\text{electrons}}\times\dfrac{1.609\times10^{-19}\,C}{1\,\cancel {electron}} \\ \text{ } \\Q=8.045\times10^{-3}\,C=8.045\,mC$

In conclusion, 50 billion million deficit of electrons will have a charge of 8.045 X 10^-3 C or 8.045 mC.

Reference:

• Sears, F. W., & Zemansky, M. W. (1973). University physics.