Question

Charge on a piece of metal is \(-3.2\) coulomb. Number of excess electrons in the metal is:

• A. \(6.25\times 10^{18}\)
• B. \(2\times 10^{19}\)
• C. \(2\times 10^{18}\)
• D. \(6.5\times 10^{16}\)

Hint:

Counting electrons? Use \(N=\dfrac{|Q|}{e}\). The sign of \(Q\) tells you electrons vs deficit; the number itself is positive.

Answer 1

The Correct Option is B

Step 1 (Quantization of charge). Electric charge on any isolated body is an integer multiple of the elementary charge \(e\) \((e=1.6\times 10^{-19}\,\text{C})\). If a body has an excess of \(N\) electrons, its charge is \(Q=-Ne\).
Step 2 (Use magnitude for counting). The negative sign in \(Q=-3.2~\text{C}\) indicates extra electrons; the count is \(N=\dfrac{|Q|}{e}\).
Step 3 (Compute). \(N=\dfrac{3.2}{1.6\times 10^{-19}}=\left(\dfrac{3.2}{1.6}\right)\times 10^{19}=2\times 10^{19}\).
(Using \(e=1.602\times 10^{-19}\,\text{C}\) gives \(N\approx 1.997\times 10^{19}\), which rounds to \(2\times 10^{19}\)).
\(\boxed{N=2\times 10^{19}\ \text{excess electrons}}\)