Question

The distance between two electrons in vacuum is 5.3 nm. Determine the work needed to reduce the distance to 1/4 of the original distance.

Answer 1

Answer:
`W=1.304* 10^(-19) J`

Step-by-step explanation:

Given

Distance between two electron is
`(r_1)5.3 nm`

Charge of electron
`=1.6* 10^(-19) C`

Final distance
`(r_2)=\farc{r_1}{4}=1.325 nm`

Work done
`=kq_1q_2\left ( (1)/(r_2)-(1)/(r_1)\right )`

`W=(1)/(4\pi \epsilon _)q_1q_2\left ( (1)/(r_2)-(1)/(r_1)\right )`

`W=9* 10^9* (1.6* 10^(-19))^2\left ( (1)/(1.325* 10^(-9))-(1)/(5.3* 10^(-9))\right )`

`W=9* 10^9* 2.56* 10^(-38)* 5.66* 10^8 J`

`W=1.304* 10^(-19) J`