Answer :
Answer:
The acceleration of an electron is [tex]1.2\times10^{20}\ m/s^2[/tex]
Explanation:
Given that,
One Charge = 50 μC
Distance on y axis = 3.0 cm
Second charge = 77 μC
Distance on x axis = 4.0 cm
We need to calculate the force on electron due to q₁
Using formula of force
[tex]F_{1}=\dfrac{kq_{1}q}{r^2}[/tex]
Here, q = charge of electron
Put the value into the formula
[tex]F_{1}=\dfrac{9\times10^{9}\times50\times10^{-6}\times1.6\times10^{-19}}{(3\times10^{-2})^2}[/tex]
[tex]F_{1}=8\times10^{-11}j\ \ N[/tex]
We need to calculate the force on electron due to q₂
Using formula of force
[tex]F_{2}=\dfrac{kq_{2}q}{r^2}[/tex]
Here, q = charge of electron
Put the value into the formula
[tex]F_{2}=\dfrac{9\times10^{9}\times77\times10^{-6}\times1.6\times10^{-19}}{(4\times10^{-2})^2}[/tex]
[tex]F_{2}=6.93\times10^{-11}i\ \ N[/tex]
We need to calculate the net force
Using formula of net force
[tex]F=F_{1}+F_{2}[/tex]
Put the value into the formula
[tex]F=8\times10^{-11}j+6.93\times10^{-11}i[/tex]
The magnitude of the net force
[tex]F=\sqrt{(8\times10^{-11})^2+(6.93\times10^{-11})^2}[/tex]
[tex]F=1.058\times10^{-10}\ N[/tex]
We need to calculate the acceleration of an electron
Using formula of force
[tex]F = ma[/tex]
[tex]a=\dfrac{F}{m}[/tex]
Put the value into the formula
[tex]a=\dfrac{1.058\times10^{-10}}{9.1\times10^{-31}}[/tex]
[tex]a=1.2\times10^{20}\ m/s^2[/tex]
Hence, The acceleration of an electron is [tex]1.2\times10^{20}\ m/s^2[/tex]