A + 3.0 nC charge Q is initially at rest at a distance of r; = 10 cm from a + 5.0 nC charge q fixed at the origin. The charge Q is moved away from q to a new position at r, = 15 cm. In this process work done by the field is

Initially when both charges are 10cm apart
Potential energy of system $={\mathrm{U}}_{\mathrm{i}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(3\mathrm{nC}\right)\left(5\mathrm{nC}\right)}{10\times {10}^{-2}\mathrm{m}}$
Finally, both charges are 15cm apart
Potential energy of system $={\mathrm{U}}_{\mathrm{j}}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(3\mathrm{nC}\right)\left(5\mathrm{nC}\right)}{15\times {10}^{-2}\mathrm{m}}$
Work done by external force $={\mathrm{W}}_{\text{ext }}={\mathrm{U}}_{\mathrm{j}}-{\mathrm{U}}_{\mathrm{i}}$
$\begin{array}{l}=\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(3\mathrm{nC}\right)\left(5\mathrm{nC}\right)}{15\times {10}^{-2}\mathrm{m}}-\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(3\mathrm{nC}\right)\left(5\mathrm{nC}\right)}{10\times {10}^{-2}\mathrm{m}}\\ =\frac{1}{4{\mathrm{\pi \epsilon }}_0}\frac{\left(3\mathrm{nC}\right)\left(5\mathrm{nC}\right)}{{10}^{-2}}\left[\frac{1}{15}-\frac{1}{10}\right]\\ =\frac{9\times {10}^9\times 3\times {10}^{-9}\times 5\times {10}^{-9}}{{10}^{-2}}\left[\frac{10-15}{150}\right]\\ =\frac{27\times {10}^{-7}[-5]}{306}=-4.5\times {10}^{-7}\mathrm{J}\end{array}$
Work done by field ${\mathrm{W}}_{\text{field }}=-{\mathrm{W}}_{\text{ext }}=-\left(-4.5\times {10}^{-7}\mathrm{J}\right)=4.5\times {10}^{-7}\mathrm{J}$