The discovery of the electron (AQA A-Level Physics): Flashcards

To determine the charge of the electron

Negatively charged due to friction

$E = \frac{V}{d}$

$EQ = mg$ or $\frac{QV}{d} = mg$

$F = 6\pi\eta r v$

$6\pi\eta r v = mg$

$r = \sqrt{\frac{9\eta v}{2\rho g}}$

Always multiples of $1.6 \times 10^{-19}$ C

$1.6 \times 10^{-19}$ C

Electric charge is quantised