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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 \( q=\lambda L \)

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