Two charges 1C and 2C are placed at coordinates (0,3) and (4,3) respectively in an XY plane. What is the work done to put a 30C charge at the origin of the coordinate system?

The potential at the origin due to the charges $q_1$ and $q_2$ is given by $V = V_1 + V_2$.

The distance of the first charge $q_1 = 1 \text{ C}$ at $(0, 3)$ from the origin is $r_1 = 3 \text{ m}$.

The distance of the second charge $q_2 = 2 \text{ C}$ at $(4, 3)$ from the origin is $r_2 = \sqrt{4^2 + 3^2} = 5 \text{ m}$.

The total electric potential at the origin is $V = \dfrac{1}{4\pi\epsilon_0} \left( \dfrac{q_1}{r_1} + \dfrac{q_2}{r_2} \right)$.

Substituting the given values, we get $V = 9 \times 10^9 \left( \dfrac{1}{3} + \dfrac{2}{5} \right)$.

$V = 9 \times 10^9 \left( \dfrac{5 + 6}{15} \right) = 9 \times 10^9 \left( \dfrac{11}{15} \right) = 6.6 \times 10^9 \text{ V}$.

The work done to bring a charge $q_3 = 30 \text{ C}$ to the origin is $W = q_3 V$.

$W = 30 \times 6.6 \times 10^9 = 198 \times 10^9 \text{ J}$.