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At time \(t=0\) a proton is a distance of 0.360 m from
a very large insulating sheet of charge and is moving parallel to the
sheet with speed \(9.70 \times 10^{2} \mathrm{m} / \mathrm{s}\) . The sheet has uniform surface
charge density \(2.34 \times 10^{-9} \mathrm{C} / \mathrm{m}^{2} .\) What is the speed of the proton
at \(t=5.00 \times 10^{-8} \mathrm{s} ?\)

\( E=\frac{6}{2 \varepsilon_{0}}=\frac{2.34*10^{-9}}{2 * 8.854 * 10^{-12}}=132.1 \mathrm{N/c} \)

\( F=e E \)

\( F=m a \longrightarrow E e=m a y \longrightarrow a y=\frac{E c}{m} \)

\(E q =E e \)

\( a y=\frac{132.1*1.602 * 10^{-19}}{1.673 * 10^{-27} \mathrm{kg}} = 1.265 * 10^{10} \mathrm{m} / \mathrm{s}^{2} \)

\( v{y}=v{ay}{+} a{y} t =\left(1.265*10^{10}\right) * 5 * 10^{-8} \mathrm{s} \)

\(∴ v y=632-7 m / s \)

\( v=\sqrt{v{x}^{2}+v y^{2}} =\sqrt{\left(9.7*10^{2}\right)^{2}+(632.7)^{2}} =1.16 * 10^{3} \mathrm{m} / \mathrm{s} \)

 

 

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