Question

When a piece of polythene is rubbed with wool, a negative charge of $4 \times 10^{-7}\text{ C}$ is developed on the polythene. The number of electrons transferred from wool to polythene is [$e = 1.6 \times 10^{-19}\text{ C}$]

• A. $1.5 \times 10^{12}$
• B. $3.5 \times 10^{13}$
• C. $2.5 \times 10^{13}$
• D. $2.5 \times 10^{12}$

Hint:

To perform the division $\frac{4}{1.6}$ quickly in your head, scale both numbers by 10 to clear the decimal point: $\frac{40}{16}$. Dividing both parts by 8 yields $\frac{5}{2} = 2.5$ in seconds!

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the total number of electrons transferred during a frictional electricity interaction. When polythene is rubbed with wool, it gains a net negative charge, indicating that electrons have moved from the wool to the polythene surface.

Step 2: Key Formula or Approach:
According to the principle of

Quantization of Charge, the total charge $q$ on a body must always be an integral multiple of the basic elementary unit of electronic charge $e$: $$q = N \cdot e$$ Where $N$ is the total number of electrons transferred, and $e = 1.6 \times 10^{-19}\text{ C}$. Rearranging the expression gives: $$N = \frac{q}{e}$$

Step 3: Detailed Explanation:
Identify the parameters provided in the problem statement: Magnitude of the total developed charge, $q = 4 \times 10^{-7}\text{ C}$ Elementary charge of an electron, $e = 1.6 \times 10^{-19}\text{ C}$
Substitute these parameters into the quantization formula to solve for $N$: $$N = \frac{4 \times 10^{-7}}{1.6 \times 10^{-19}}$$ Simplify the expression by handling the decimal coefficients and exponents separately: $$N = \frac{4}{1.6} \times 10^{-7 - (-19)}$$ $$N = 2.5 \times 10^{12}$$ This shows that exactly $2.5 \times 10^{12}$ electrons were transferred from the wool to the polythene sheet, which matches option (D).

Step 4: Final Answer:
The number of transferred electrons is $2.5 \times 10^{12}$, corresponding to option (D).