Question

A 1 μC point charge is held at the origin of a cartesian coordinate system. If a second point charge of 10 μC is moved from (0, 10, 0) to (5, 5, 5) and subsequently to (5, 0, 0), then the total work done is _________ mJ.

(Round off to 2 decimal places).

Take \(\frac {1}{4\pi \varepsilon_o} = 9 \times 10^9\) in SI units. All coordinates are in meters.

Answer 1

Concept:

Electric Potential: It is the amount of work needed to move a unit charge from a reference point to a specific point against the electric field.

If the two points are specified then the total work done in moving the charge from one point to another with respect to the reference point charge is given as,

\(W=-\frac{1}{4\pi {{\epsilon }_{0}}}{{q}_{1}}{{q}_{2}}\left( \frac{1}{{{r}_{1}}}-\frac{1}{{{r}_{2}}} \right)=k{{q}_{1}}{{q}_{2}}\left( \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right)\)

Where, k = 9 × 109 N-m2 C-2

q1 and q2 = charges

r1 and r2 are distances

Electric Potential due to point charge: Consider a point charge is placed at the distance of ‘r’ from the charge ‘Q’ then the potential due to charge is,

\({{V}_{p}}=\frac{kQ}{r}\)

Calculation:

Given, q₁ = 1 μC, q₂ = 10 μC

Let r₁ be the distance the two-point charges one placed placed at (0, 0, 0) and other at (0, 10 , 0)

r₁ = 10 m

Let r₂ be the final distance between the two-point charges one placed at origin (0, 0, 0) and other at (5, 0, 0)

r₂ = 5 m

Total work done in moving the pont charge from (0, 0, 0) to (5, 0, 0) is

\(W=k{{q}_{1}}{{q}_{2}}\left( \frac{1}{{{r}_{2}}}-\frac{1}{{{r}_{1}}} \right)\)

\(W=9\times 10^9\times 10^{-11}\left( \frac{1}{5}-\frac{1}{10} \right)\)

W = 9 mJ