Find work done in moving a 2 μC charge from A to B.

Given charge Z₁ = 10^-8 C

The correct answer is 34.3 μJ.

1. Understand the Geometry and Given Values:
Based on the provided illustration, we have:
- A source charge $Z_1={10}^{-8}\text{C}$ located at the origin $(0,0,0)$.
- Point A located at the coordinates $(5,4,2\sqrt{2})$.
- Point B located at the coordinates $(2,2,1)$.
- The test charge to be moved is $q=2\mu \text{C}=2\times {10}^{-6}\text{C}$ from A to B.

2. Calculate the Distances from the Origin:
First, find the distance of point A from the source charge at the origin:
$r_A=\sqrt{5^2+4^2+{\left(2\sqrt{2}\right)}^2}=\sqrt{25+16+8}=\sqrt{49}=7\text{m}$
Next, find the distance of point B from the source charge at the origin:
$r_B=\sqrt{2^2+2^2+1^2}=\sqrt{4+4+1}=\sqrt{9}=3\text{m}$

3. Formula for Work Done:
The work done in moving a test charge $q$ from potential $V_A$ to potential $V_B$ is given by the difference in potential energy:
$W=q(V_B-V_A)$
The electrostatic potential due to a point charge is $V=\frac{kQ}{r}$, which gives:
$W=q·k{Z}_1(\frac{1}{r_B}-\frac{1}{r_A})$
where Coulomb's constant $k=9\times {10}^9\text{N}·{\text{m}}^2/{\text{C}}^2$.

4. Compute the Value:
Substitute the given values into the equation:
$W=(2\times {10}^{-6})\times (9\times {10}^9)\times {10}^{-8}\times (\frac{1}{3}-\frac{1}{7})$
Simplify the constants outside the brackets:
$W=18\times {10}^{-5}\times \left(\frac{7-3}{21}\right)$
$W=18\times {10}^{-5}\times \frac{4}{21}\text{J}$
$W=\frac{72}{21}\times {10}^{-5}\text{J}\approx 3.43\times {10}^{-5}\text{J}=34.3\mu \text{J}$