a) [tex]2.88\cdot 10^{11} N/C[/tex]
b) [tex]1.44\cdot 10^{11} N/C[/tex]
c) [tex]4.6\cdot 10^{-8} N[/tex]
Explanation:
a)
The magnitude of the electric field produced by a single-point charge is given by the equation
[tex]E=\frac{kQ}{r^2}[/tex]
where
k is the Coulomb's constant
Q is the charge source of the field
r is the distance from the charge
The field has direction:
- Out from the charge if the charge is positive
- Towards the charge if the charge is negative
In this part, we want to find the electric field due to the alpha particle at the location of the electron. We have:
[tex]q=3.2\cdot 10^{-19}C[/tex] (charge of the alpha particle)
[tex]r=10^{-10}m[/tex] distance of the electron from the alpha particle
So, the electric field is
[tex]E=\frac{(9\cdot 10^9)(3.2\cdot 10^{-19})}{(10^{-10})^2}=2.88\cdot 10^{11} N/C[/tex]
And since the alpha particle is positively charged, the direction of the field is out from the particle.
b)
Here we want to find the electric field due to the electron at the location of the alpha particle.
We use again the same formula as before:
[tex]E=\frac{kQ}{r^2}[/tex]
where
k is the Coulomb's constant
Q is the charge source of the field
r is the distance from the charge
where in this case:
[tex]Q=1.6\cdot 10^{-19}C[/tex] is the charge of the electron
[tex]r=10^{-10}m[/tex] distance of the alpha particle from the electron
Therefore, the electric field due to the electron is
[tex]E=\frac{(9\cdot 10^9)(1.6\cdot 10^{-19})}{(10^{-10})^2}=1.44\cdot 10^{11} N/C[/tex]
And since the electron is negatively charged, the direction of the field is towards the electron.
c)
The electrostatic force between two electric charges is given by Coulomb's Law:
[tex]F=\frac{kq_1 q_2}{r^2}[/tex]
where
k is the Coulomb's constant
q1, q2 are the two charges
r is the separation between the two charges
In this problem we have:
[tex]q_1=3.2\cdot 10^{-19}C[/tex] is the charge of the alpha particle
[tex]q_2=1.6\cdot 10^{-19}C[/tex] is the charge of the electron
[tex]r=10^{-10}m[/tex] distance between the two charges
So, the force is
[tex]F=\frac{(9\cdot 10^9)(3.2\cdot 10^{-19})(1.6\cdot 10^{-19})}{(10^{-10})^2}=4.6\cdot 10^{-8} N[/tex]
The force is the same on the two particles (Newton's third law of motion), and it is attractive, since the two charges have opposite sign.
