Two point charges of -5 μ C and 2 μ C are located in free space at ( -4 cm, 0 ) and ( 6 cm, 0 ) respectively. (a) Calculate the amount of work done…

(a) \(\displaystyle \mathrm{U}=\frac{\mathrm{kq}_{1} \mathrm{q}_{2}}{\mathrm{r}}\) \[\begin{aligned} & =\frac{9 \times 10^{9} \times\left(-5 \times 10^{-6}\right) \times\left(2 \times 10^{-6}\right)}{10 \times 10^{-2}} \\ & =-0.9 \mathrm{~J} \end{aligned} \] \[\mathrm{W}=\mathrm{U}(\text { infinity })-\mathrm{U}(\mathrm{r})=0.9 \mathrm{~J} \]

(b) \(\displaystyle \mathrm{E}=\frac{-\mathrm{dV}}{\mathrm{dr}}\) \[\begin{aligned} & d V=\frac{-\mathrm{A}}{\mathrm{r}^{2}} d \mathrm{r} \\ & V=\frac{\mathrm{A}}{\mathrm{r}} \end{aligned} \] \[\mathrm{U}=\mathrm{q}_{1} V_{1}+\mathrm{q}_{2} V_{2}+\frac{k \mathrm{q}_{1} \mathrm{q}_{2}}{\mathrm{r}} \] \[\begin{aligned} = & \left(-5 \times 10^{-6}\right)\left(\frac{8 \times 10^{4}}{4 \times 10^{-2}}\right)+\left(2 \times 10^{-6}\right)\left(\frac{8 \times 10^{4}}{6 \times 10^{-2}}\right)-0.9 \\ & =-10+2.67-0.9 \\ & =-8.24 \mathrm{~J} \end{aligned} \]