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:$W=1.304\times 10^{-19} J$

Explanation:

Given

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

Charge of electron$=1.6\times 10^{-19} C$

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

Work done$=kq_1q_2\left ( \frac{1}{r_2}-\frac{1}{r_1}\right )$

$W=\frac{1}{4\pi \epsilon _}q_1q_2\left ( \frac{1}{r_2}-\frac{1}{r_1}\right )$

$W=9\times 10^9\times (1.6\times 10^{-19})^2\left ( \frac{1}{1.325\times 10^{-9}}-\frac{1}{5.3\times 10^{-9}}\right )$

$W=9\times 10^9\times 2.56\times 10^{-38}\times 5.66\times 10^8 J$

$W=1.304\times 10^{-19} J$