Question

A 1 µC point charge is held at the origin. A second point charge of 10 µC is moved from (0,10,0) to (5,5,5) and then to (5,0,0). The total work done is \(\underline{\hspace{2cm}}\) mJ. (Round off to 2 decimal places).
Take \[ \frac{1}{4\pi\epsilon_0} = 9 \times 10^{9} \]

Hint:

Work done in moving charges equals change in electric potential energy, independent of path.

Answer 1

Correct Answer: 8.9

The work done in moving a charge in an electric field is equal to the change in electric potential energy.
Let the fixed charge be: \[ q_1 = 1\,\mu C = 1 \times 10^{-6} C \] Moving charge: \[ q_2 = 10\,\mu C = 10 \times 10^{-6} C \] Distance from origin at three positions:
Initial position: \[ r_1 = 10 \] Intermediate position: \[ r_2 = \sqrt{5^2 + 5^2 + 5^2} = \sqrt{75} = 8.66 \] Final position: \[ r_3 = 5 \] Potential energies: \[ U = k \frac{q_1 q_2}{r} \] Compute values:
\[ U_1 = 9\times10^{9}\frac{(1 \times 10^{-6})(10 \times10^{-6})}{10} = 0.009\text{ J} \] \[ U_2 = 9\times10^{9}\frac{10^{-11}}{8.66} = 0.01038\text{ J} \] \[ U_3 = 9\times10^{9}\frac{10^{-11}}{5} = 0.018\text{ J} \] Total work done: \[ W = U_3 - U_1 = 0.018 - 0.009 = 0.009\text{ J} \] \[ \boxed{W = 9.00\ \text{mJ}} \]