Answer:
The value is [tex]a =1.233 *10^{20} \ m/s^2[/tex]
Explanation:
From the question we are told that
The magnitude of the first charge is [tex]q_1 = 31 \mu C = 31 *10^{-6} \ C[/tex]
The position is y = 8 cm = 0.08 m
The magnitude of the second charge is [tex]q_1 = 55 \mu C = 55 *10^{-6} \ C[/tex]
The position is x = 3 cm = 0.03 m
Generally the force exerted on the electron by the first charge is mathematically represented as
[tex]F_1 = \frac{k * q_1 * e }{y^2}[/tex]
Here k is the coulombs constants with value
=> [tex] k =9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.[/tex]
and e is the charge on a electron with value [tex]e = 1.60 *10^{-19} \ C[/tex]
So
[tex]F_1 = \frac{9*10^{9} * 31 *10^{-6} * 1.60 *10^{-19} }{0.08^2}[/tex]
[tex]F_1 = 6.975 *10^{-11} j \ N[/tex]
Generally the force exerted on the electron by the first charge is mathematically represented as
[tex]F_2 = \frac{k * q_2 * e }{x^2}[/tex]
=> [tex]F_2 = \frac{9*10^{9} * 55 *10^{-6} * 1.60 *10^{-19} }{0.03^2}[/tex]
=> [tex]F_2 = 8.8*10^{-11} i \ N[/tex]
Generally the net force exerted is mathematically represented as
[tex]F_n = F_1 + F_2[/tex]
So
[tex]F_n = 6.975 *10^{-11} j + 8.8*10^{-11} i [/tex]
The resultant of this net force is mathematically represented as
[tex]F_nr = \sqrt{ (6.975 *10^{-11})^2 + (8.8*10^{-11})^2}[/tex]
[tex]F_nr = \sqrt{4.865*10^{-21} + 7.74*10^{-21}}[/tex]
[tex]F_nr = 1.123 *10^{-10}\ N [/tex]
Generally this force can be represented as
[tex]F_nr = ma[/tex]
Here m is the mass of the electron with value
[tex]m = 9.11 *10^{-31} \ kg[/tex]
[tex]a = \frac{F_{nr}}{m}[/tex]
=> [tex]a = \frac{1.123 *10^{-10}}{9.11 *10^{-31}}[/tex]
=> [tex]a =1.233 *10^{20} \ m/s^2[/tex]
