• عربي

Need Help?

Subscribe to Physics 2

###### \${selected_topic_name}
• Notes

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 $t=5.00 \times 10^{-8} \mathrm{s} ?$

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

$F=e E$

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

$E q =E e$

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

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

$∴ v y=632-7 m / s$

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