Question

An ebonite rod acquires a negative change of $6.4\times {10}^{-10}$C. The number of excess electrons it has is

• A. $4\times {10}^9$
• B. $4\times {10}^{-9}$
• C. $4\times {10}^{-29}$
• D. $4\times {10}^{29}$

Answer 1

Answer: (A)

$4\times {10}^9$

If $n$ be the number of excess electrons, then net charge $q=ne$.

Thus, $n=\frac{q}{e}=\frac{-6.4\times {10}^{-10}}{-1.6\times {10}^{-19}}=4\times {10}^9$