Respuesta :
Answer:
Explanation:
The work done on alpha is given by
Charge on alpha particle is 2 the electron charge
[tex]W=\Delta U=U_{b}-U_{a} \quad \Rightarrow(1)[/tex]
The potential at center [tex]\left[U_{a}\right][/tex]
The electric potential due to multiples point charges is given by
[tex]U_{a}=k q q_{a} \sum_{i=1}^{4}\left(\frac{1}{r_{i}}\right) \quad \Rightarrow(2)[/tex]
where [tex]q[/tex]=charge on electron
[tex]q_{\alpha }[/tex]=charge on alpha particle
But and is equal for all charges so ( 2) is equal to
[tex]U_{a}=\frac{4 k(-e)(2 e)}{d} \quad \Rightarrow(3)[/tex]
The radius is given by
[tex]d=\frac{1}{2} \sqrt{(10)^{2}+(10)^{2}}=7.071 \mathrm{nC}[/tex]
Substitution in (3) yields
[tex]U_{a}=\frac{-8 \times 9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{7.071 \times 10^{-9}}=-2.607 \times 10^{-19} \mathrm{J}[/tex]
The potential at mid point [tex]\left[U_{b}\right][/tex]
due to lower charges the radius is given by
[tex]r_{t}=\sqrt{L^{2}+(L / 2)^{2}}=\sqrt{10^{2}+5^{2}}=11.18 \mathrm{nm}[/tex]
And for upper charges the radius is
[tex]r_{u}=5 \mathrm{nm}[/tex]
So the electric potential is
[tex]U_{b} &=k q q_{a}\left[\frac{2}{r_{l}}+\frac{2}{r_{u}}\right][/tex]
[tex]=k(-e)(2 e)\left[\frac{2}{r_{1}}+\frac{2}{r_{u}}\right][/tex]
[tex]=-4 \times\left(1.6 \times 10^{9}\right)\left(1.6 \times 10^{-19}\right)^{2}\left[\frac{1}{11.18 \times 10^{-9}}+\frac{1}{5.0 \times 10^{-9}}\right][/tex]
[tex]=-2.667 \times 10^{-19} \mathrm{J}[/tex]
The work done on the charge is equal to
[tex]W=U_{b}-U_{a}=-2.667 \times 10^{-19}+2.607 \times 10^{-19}=-6.00 \times 10^{-21} \mathrm{J}[/tex]
The work done on is given by
[tex]W=\Delta U=U_{b}-U_{a} \quad \Rightarrow(1)[/tex]
The potential at center [tex]\left[U_{a}\right][/tex]
The electric potential due to multiples point charges is given by
[tex]U_{a}=k q q_{a} \sum_{i=1}^{4}\left(\frac{1}{r_{i}}\right) \quad \Rightarrow(2)[/tex]
But and is equal for all charges so ( 2) is equal to
[tex]U_{a}=\frac{4 k(-e)(2 e)}{d} \quad \Rightarrow(3)[/tex]
The radius is given by
[tex]d=\frac{1}{2} \sqrt{(10)^{2}+(10)^{2}}=7.071 \mathrm{nC}[/tex]
Substitution in (3) yields
[tex]U_{a}=\frac{-8 \times 9 \times 10^{9} \times\left(1.6 \times 10^{-19}\right)^{2}}{7.071 \times 10^{-9}}=-2.607 \times 10^{-19} \mathrm{J}[/tex]
The potential at mid point [tex]\left[U_{b}\right][/tex]
due to lower charges the radius is given by
[tex]r_{t}=\sqrt{L^{2}+(L / 2)^{2}}=\sqrt{10^{2}+5^{2}}=11.18 \mathrm{nm}[/tex]
And for upper charges the radius is
[tex]r_{u}=5 \mathrm{nm}[/tex]
So the electric potential is
[tex]U_{b} &=k q q_{a}\left[\frac{2}{r_{l}}+\frac{2}{r_{u}}\right][/tex]
[tex]=k(-e)(2 e)\left[\frac{2}{r_{1}}+\frac{2}{r_{u}}\right][/tex]
[tex]=-4 \times\left(1.6 \times 10^{9}\right)\left(1.6 \times 10^{-19}\right)^{2}\left[\frac{1}{11.18 \times 10^{-9}}+\frac{1}{5.0 \times 10^{-9}}\right][/tex]
[tex]=-2.667 \times 10^{-19} \mathrm{J}[/tex]
The work done on the charge is equal to
[tex]W=U_{b}-U_{a}=-2.667 \times 10^{-19}+2.607 \times 10^{-19}=-6.00 \times 10^{-21} \mathrm{J}[/tex]
[tex]W=-6.00 \times 10^{-21} \mathrm{J}[/tex]
