Find out work done in moving a $2\,\mu$C charge from point $A$ to $B$.
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In electrostatics, work done depends only on initial and final positions. Always calculate distances from the source charge and use potential difference instead of force integration.
Concept:
Work done in moving a charge in an electrostatic field depends only on the initial and final positions, not on the path followed. It is given by:
\[
W = q\,(V_A - V_B)
\]
where electric potential due to a point charge is:
\[
V = \frac{1}{4\pi\varepsilon_0}\frac{Q}{r}
\]
Step 1: Calculate distances of points from the origin
For point $A(5,4,2\sqrt{2})$:
\[
r_A = \sqrt{5^2 + 4^2 + (2\sqrt{2})^2}
= \sqrt{25 + 16 + 8}
= \sqrt{49} = 7
\]
For point $B(2,2,1)$:
\[
r_B = \sqrt{2^2 + 2^2 + 1^2}
= \sqrt{9} = 3
\]
Step 2: Write expression for work done
\[
W = q \left( \frac{1}{4\pi\varepsilon_0} Q \left( \frac{1}{r_A} - \frac{1}{r_B} \right) \right)
\]
Step 3: Substitute values
\[
q = 2 \times 10^{-6}\text{ C}, \quad Q = 10^{-8}\text{ C}
\]
\[
W = (2 \times 10^{-6})(9 \times 10^9)(10^{-8})
\left( \frac{1}{7} - \frac{1}{3} \right)
\]
Step 4: Simplify
\[
W = 1.8 \times 10^{-4} \left( \frac{3 - 7}{21} \right)
= -3.43 \times 10^{-5} \text{ J}
\]
Magnitude of work done:
\[
|W| = 34.3\,\mu\text{J}
\]
Conclusion:
\[
\boxed{W = 34.3\,\mu\text{J}}
\]
