First we confirm how many electrons were removed:
N=50 billion million electronsN=(50)(109)(106) electronsN=5×1016 electrons
Since we are removing electrons (because this is a deficit), the total charge will be positive:
Q=5×1016electrons×1electron1.609×10−19C Q=8.045×10−3C=8.045mC
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.
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