Work Done Calculation for Point Charges

The problem asks for the work done ($W$) in bringing a test charge ($q_3$) from infinity to a specific point. This work is equal to the change in potential energy, which can be calculated as the charge being moved multiplied by the potential difference.

Key Concepts:

• Work Done ($W$) = $q_3 \times (V_{\text{final}} - V_{\text{initial}})$
• Potential at infinity ($V_{\text{initial}}$) is 0.
• Potential ($V$) at a point due to a point charge ($q$) at distance ($r$) is $V = k \frac{q}{r}$, where $k = \frac{1}{4\pi\epsilon_0}$.
• Total potential is the algebraic sum of potentials due to individual charges.

Step-by-Step Solution

1. Identify Given Values:
   • Charge 1: $q_1 = 1 \text{ nC} = 1 \times 10^{-9} \text{ C}$
   • Charge 2: $q_2 = 2 \text{ nC} = 2 \times 10^{-9} \text{ C}$
   • Test Charge: $q_3 = 3 \text{ nC} = 3 \times 10^{-9} \text{ C}$
   • Side of equilateral triangle: $a = 3 \text{ cm} = 0.03 \text{ m}$
   • Coulomb's constant: $k = 9 \times 10^9 \text{ N.m}^2/\text{C}^2$
2. Calculate Potential at the Third Corner:

   The third corner is equidistant ($a$) from both $q_1$ and $q_2$. The total potential ($V$) at the third corner is the sum of the potentials due to $q_1$ and $q_2$. $V = V_1 + V_2 = k \frac{q_1}{a} + k \frac{q_2}{a}$ $V = \frac{k}{a} (q_1 + q_2)$

   Substitute the values:

   $V = \frac{9 \times 10^9 \text{ N.m}^2/\text{C}^2}{0.03 \text{ m}} (1 \times 10^{-9} \text{ C} + 2 \times 10^{-9} \text{ C})$ $V = \frac{9 \times 10^9}{0.03} (3 \times 10^{-9})$ $V = \frac{27}{0.03} = 900 \text{ V}$
3. Calculate Work Done:

   The work done to bring charge $q_3$ from infinity (where potential is 0) to the third corner (where potential is $V$) is:

   $W = q_3 \times V$ $W = (3 \times 10^{-9} \text{ C}) \times (900 \text{ V})$ $W = 2700 \times 10^{-9} \text{ J}$ $W = 2.7 \times 10^{-6} \text{ J}$
4. Convert to Microjoules ($\mu$J):

   Since $1 \mu\text{J} = 10^{-6} \text{ J}$, the work done is:

   $W = 2.7 \mu\text{J}$