Respuesta :
Answer:
[tex]4.4443704375\times 10^{-18}\ C/m^2[/tex]
Explanation:
[tex]\epsilon_0[/tex] = Permittivity of free space = [tex]8.85\times 10^{-12}\ F/m[/tex]
[tex]\Delta l[/tex] = Distance charge traveled = 2 m
v = Velocity of electron = 420 m/s
E = Electric field
[tex]m_e[/tex] = Mass of electron = [tex]9.11\times 10^{-31}\ kg[/tex]
[tex]q_e[/tex] = Charge of electron = [tex]1.6\times 10^{-19}\ C[/tex]
As the energy of the system is conserved we have
[tex]q_eE\Delta l=\dfrac{1}{2}m_ev^2\\\Rightarrow E=\dfrac{1}{2}\dfrac{m_e}{q_e}\times \dfrac{v^2}{\Delta l}\\\Rightarrow E=\dfrac{1}{2}\dfrac{9.11\times 10^{-31}}{1.6\times 10^{-19}}\times \dfrac{420^2}{2}\\\Rightarrow E=2.51094375\times 10^{-7}\ N/C[/tex]
For an infinite non conducting sheet electric field is given by
[tex]E=\dfrac{\sigma}{2\epsilon}\\\Rightarrow \sigma=2E\epsilon\\\Rightarrow \sigma=2\times 2.51094375\times 10^{-7}\times 8.85\times 10^{-12}\\\Rightarrow \sigma=4.4443704375\times 10^{-18}\ C/m^2[/tex]
The surface charge density is [tex]4.4443704375\times 10^{-18}\ C/m^2[/tex]
The surface charge density on the sheet is 1.77 x 10⁻¹⁷ C/m².
The given parameters;
- initial position of the electron, x₁ = 3 m
- initial speed of the electron, v = 420 m/s
- final position of the electron, x₂ = 2m
The electric field on the charged electron is calculated from the principle of conservation of energy;
[tex]F\Delta d= \frac{1}{2} mv^2\\\\Eq \Delta d = \frac{1}{2} mv^2\\\\E = \frac{mv^2}{q \times \Delta d} \\\\E = \frac{9.11 \times 10^{-31} \times 420^2}{1.602 \times 10^{-19} \times (2-1)} \\\\E = 1. 00 \times 10^{-6} \ N/C[/tex]
The charge density is calculated as follows;
[tex]E = \frac{\sigma }{2 \epsilon} \\\\\sigma = E \times 2\epsilon \\\\\sigma = 1.00 \times 10^{-6} \times 2 \times (8.85 \times 10^{-12} )\\\\\sigma = 1.77 \times 10^{-17} \ C/m^2[/tex]
Thus, the surface charge density on the sheet is 1.77 x 10⁻¹⁷ C/m².
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