Question

A charge of 31 micro-C is placed on the y axis at y = 8 cm and a 55 micro-C charge is placed on the x axis at x = 3 cm. If both charges are held fixed, what is the magnitude of the initial acceleration of an electron released from rest at the origin? Write your answer in terms of 10^18 m/s2.

Answer 1

Answer:

The value is $a =1.233 *10^{20} \ m/s^2$

Explanation:

From the question we are told that

The magnitude of the first charge is $q_1 = 31 \mu C = 31 *10^{-6} \ C$

The position is y = 8 cm = 0.08 m

The magnitude of the second charge is $q_1 = 55 \mu C = 55 *10^{-6} \ C$

The position is x = 3 cm = 0.03 m

Generally the force exerted on the electron by the first charge is mathematically represented as

$F_1 = \frac{k * q_1 * e }{y^2}$

Here k is the coulombs constants with value

=> $k =9*10^{9}\ kg\cdot m^3\cdot s^{-4} \cdot A^{-2}.$

and e is the charge on a electron with value $e = 1.60 *10^{-19} \ C$

So

$F_1 = \frac{9*10^{9} * 31 *10^{-6} * 1.60 *10^{-19} }{0.08^2}$

$F_1 = 6.975 *10^{-11} j \ N$

Generally the force exerted on the electron by the first charge is mathematically represented as

$F_2 = \frac{k * q_2 * e }{x^2}$

=> $F_2 = \frac{9*10^{9} * 55 *10^{-6} * 1.60 *10^{-19} }{0.03^2}$

=> $F_2 = 8.8*10^{-11} i \ N$

Generally the net force exerted is mathematically represented as

$F_n = F_1 + F_2$

So

$F_n = 6.975 *10^{-11} j + 8.8*10^{-11} i$

The resultant of this net force is mathematically represented as

$F_nr = \sqrt{ (6.975 *10^{-11})^2 + (8.8*10^{-11})^2}$

$F_nr = \sqrt{4.865*10^{-21} + 7.74*10^{-21}}$

$F_nr = 1.123 *10^{-10}\ N$

Generally this force can be represented as

$F_nr = ma$

Here m is the mass of the electron with value

$m = 9.11 *10^{-31} \ kg$

       $a = \frac{F_{nr}}{m}$

=>       $a = \frac{1.123 *10^{-10}}{9.11 *10^{-31}}$

=> $a =1.233 *10^{20} \ m/s^2$