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

The value is
`a =1.233 *10^(20) \ m/s^2`

Step-by-step 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 = (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 = (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 = (k * q_2 * e )/(x^2)`

=>
`F_2 = (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 =  (F_(nr))/(m)`

=>
`a =  (1.123 *10^(-10))/(9.11 *10^(-31))`

=>
`a =1.233 *10^(20) \ m/s^2`