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• Notes

How many excess electrons must be added to an isolated
spherical conductor 32.0 cm in diameter to produce an electric
field of 1150 N / C  just outside the surface?

$E=\frac{1}{4 \pi \varepsilon_{0}} * \frac{q}{R^{2}}$

$∴ q=\frac{R^{2} E}{1/4 \pi \varepsilon_{0}}= \frac{(0.16)^{2} * 1150}{9+10^{9}}=3.275 * 10^{-9} \mathrm{C}$

$∵ e=1.602 * 10^{-19} \mathrm{c}$

#excess elect. $=\frac{3.275 * 10^{-9}}{1.602*10^{-19}}=2.04 * 10^{10}$ elect.

The electric field 0.400 m from a very long uniform line
of charge is 840 N / C . How much charge is contained in a $2.00-\mathrm{cm}$
section of the line?

$\oint \vec{f} \cdot \overrightarrow{d A}= E A_{\text { cylinder }}= E(2 \pi r L)$

$=840 *(2 \pi * 0.4 * 0.02)=42.2 \mathrm{N.m}^{2} / \mathrm{c}$

$\phi_{\varepsilon}=42.2 \quad N \cdot m^{2} / c$

$q=\varepsilon_{0} \times \phi_{\varepsilon}=8.854+10^{-12} * 42.2=3.74 * 10^{-10} \mathrm{C}$

OR $q=\lambda L$