Question

An ebonite rod acquires a negative charge of $3.2\times {10}^{-10}$C. The number of excess electrons it has is given by

• A. $2\times {10}^9$
• B. $2\times {10}^{-9}$
• C. $2\times {10}^{-29}$
• D. $2\times {10}^{29}$

Answer 1

The correct option is A $2\times {10}^9$

Total charge $q=ne$

Thus, $n=\frac{q}{e}=\frac{-3.2\times {10}^{-10}}{1.6\times {10}^{-19}}=2\times {10}^9$