Question

A polythene piece rubbed with wool is found to have a negative charge of $3\times10^{-7}\,\text{C}$.
(a) Estimate the number of electrons transferred (from which to which?).
(b) Is there a transfer of mass from wool to polythene?

Hint:

Polythene is negative, so it gains electrons from wool. Use \(N = |q|/e\), then mass transferred \(\Delta m = N\,m_e\).

Answer 1

Given: charge on polythene \(q = -3\times10^{-7}\,\text{C}\); charge of one electron \(e = 1.6\times10^{-19}\,\text{C}\); electron mass \(m_e = 9.1\times10^{-31}\,\text{kg}\).

Step 1: Direction of transfer. The polythene becomes negatively charged. A body becomes negative by gaining electrons. So electrons are transferred from the wool to the polythene.

Step 2: Charge quantisation. Charge comes in whole multiples of \(e\): \(q = N e\). Hence the number of electrons transferred is

\[ N = \frac{|q|}{e}. \]

Step 3: Substitute and compute.

\[ N = \frac{3\times10^{-7}}{1.6\times10^{-19}}. \]

\[ N = 1.875\times10^{12}. \]

So about \(1.9\times10^{12}\) electrons are transferred from wool to polythene.

Step 4: Mass transfer (part b). Yes. Electrons carry mass, so when electrons move from wool to polythene, mass also moves from wool to polythene. The mass transferred is

\[ \Delta m = N\,m_e = (1.875\times10^{12})(9.1\times10^{-31}). \]

\[ \Delta m = 1.706\times10^{-18}\,\text{kg}. \]

This is about \(1.7\times10^{-18}\,\text{kg}\), an extremely tiny (negligible) amount, but it is non-zero.

\[\boxed{N = 1.875\times10^{12}, \quad \Delta m \approx 1.7\times10^{-18}\,\text{kg (wool to polythene)}}\]

Answer 2

Expert framing: conservation of charge plus the electron as the charge and mass carrier.

Step 1: Rubbing does not create charge; it merely moves electrons from one surface to the other (conservation of charge). The polythene ends up with \(q = -3\times10^{-7}\,\text{C}\), so it has an excess of electrons, while the wool is left with an equal positive charge. The electrons therefore flowed wool \(\to\) polythene.

Step 2: Each transferred electron carries charge \(-e\). If \(N\) electrons move, the polythene's charge is \(-Ne\). Matching magnitudes:

\[ Ne = 3\times10^{-7}\ \Rightarrow\ N = \frac{3\times10^{-7}}{1.6\times10^{-19}} = 1.875\times10^{12}. \]

Step 3 (mass via a ratio argument): Mass and charge are carried together by the same electrons, so the transferred mass per unit transferred charge is fixed at \(m_e/e\). Thus

\[ \Delta m = \frac{m_e}{e}\,|q| = \frac{9.1\times10^{-31}}{1.6\times10^{-19}}\times(3\times10^{-7}). \]

\[ \Delta m = (5.6875\times10^{-12})(3\times10^{-7}) = 1.706\times10^{-18}\,\text{kg}. \]

Step 4: Identical result, \(1.7\times10^{-18}\,\text{kg}\), confirming the mass moves from wool to polythene (the direction the electrons go). It is utterly negligible compared with the mass of the polythene, which is why rubbing seems not to change an object's weight.

\[\boxed{N = 1.875\times10^{12}, \quad \Delta m \approx 1.7\times10^{-18}\,\text{kg}}\]