Answer to Question #98732 in Electricity and Magnetism for bryan

Question

A +7.5-nC point charge is 5.0 cm from a -9.4-μC point charge in your laboratory in
California. How much work would you have to do if you left the +7.5-nC charge in the lab
but took the -9.4-μC charge to New York City?

Answer 1

distance between charges in California

$r_1=5\cdot 10^{-2} m$

distance between charges after moving

$r_2=3.919,91km=3,91991\cdot 10^{6} m$

$r_2$ - distance between Lfliforia and New York

Then work is equal

$A=\intop_{r_1}^{r_2}E\cdot g_- \cdot dr=\frac{g_- \cdot g_+}{4 \pi \cdot \epsilon_0 }(\frac{1}{r_1}-\frac{1}{r_2})=$

$=\frac{-9.4 \cdot10^{-6}(C) \cdot 7.5 \cdot10^{-9}(C)}{4 \pi \cdot 8.85 \cdot10^{-12}(C^2/N\cdot m^2)}(\frac{1}{5\cdot 10^{-2} (m)}-\frac{1}{,91991\cdot 10^{6}(m)})=-0.013(J)$

A minus sign indicates that work is performed against electric field forces