Question

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

Required:

What is the speed of the electron when it is 10.0 cm from the +3.00-nC charge?

Answer 1

Complete Question

Two stationary positive point charges, charge 1 of magnitude 3.25 nC and charge 2 of magnitude 2.00 nC , are separated by a distance of 58.0 cm . An electron is released from rest at the point midway between the two charges, and it moves along the line connecting the two charges.

Required:

What is the speed of the electron when it is 10.0 cm from the +3.25-nC charge?

Answer:

The velocity is
`v = 80.82 \ m/s`

Step-by-step explanation:

From the question we are told that

The magnitude of charge one is
`q_1 = 3.25 nC = 3.25 *10^(-9) \ C`

The magnitude of charge two
`q_2 = 2.00 \ nC = 2.00 *10^(-9) \ C`

The distance of separation is
`d = 58.0 \ cm = 0.58 \ m`

Generally the electric potential of the electron at the midway point is mathematically represented as

`V = ( q_1 )/((d)/(2) ) + ( q_2)/((d)/(2) )`

substituting values

`V = ( 3.25 *10^(-9) )/(( 0.58)/(2) ) + ( 2 *10^(-9) )/(( 0.58)/(2) )`

`V = 1.8103 *10^(-8) \ V`

Now when the electron is 10 cm = 0.10 m from charge 1 , it is (0.58 - 0.10 = 0.48 m ) m from charge two

Now the electric potential at that point is mathematically represented as

`V_1 = (q_1)/( 0.10) + (q_2)/( 0.48)`

substituting values

`V_1 = (3.25 *10^(-9))/( 0.10) + (2.0*10^(-9))/( 0.48)`

`V_1 = 3.67*10^(-8) \ V`

Now the law of energy conservation ,

The kinetic energy of the electron = potential energy of the electron

i.e
`(1)/(2) * m * v^2 = [V_1 - V]* q`

where q is the magnitude of the charge on the electron with value

`q = 1.60 *10^(-19) \ C`

While m is the mass of the electron with value
`m = 9.11*10^(-31) \ kg`

`(1)/(2) * 9.11 *10^(-19) * v^2 = [ (3.67 - 1.8103) *10^(-8)]* 1.60 *10^(-19)`

`v = √(6532.4)`

`v = 80.82 \ m/s`