Question

The work done to keep three charges \( 2 \times 10^{-5} \text{ C} \), \( 3 \times 10^{-5} \text{ C} \), \( 4 \times 10^{-5} \text{ C} \) at vertices of an equilateral triangle of side 10 cm is:

• A. \( 324 \text{ J} \)
• B. \( 234 \text{ J} \)
• C. \( 432 \text{ J} \)
• D. \( 224 \text{ J} \)

Hint:

When calculating work done in assembling charges, ensure the distance \( r \) is in meters.

Answer 1

The Correct Option is B

Step 1: Work equals stored potential energy.
The work to assemble fixed charges equals the total electrostatic energy, which is the sum over each unique pair: \[ W = k\left(\frac{q_1q_2}{r} + \frac{q_2q_3}{r} + \frac{q_3q_1}{r}\right). \]
Step 2: All pair distances are equal.
On an equilateral triangle of side $10\,\text{cm}$, every pair is separated by $r = 0.1\,\text{m}$, so $r$ factors out.
Step 3: Take out the common factor.
\[ W = \frac{k}{r}\,(q_1q_2 + q_2q_3 + q_3q_1). \]
Step 4: Compute the charge products.
With charges $2, 3, 4$ (in units of $10^{-5}\,\text{C}$): $q_1q_2 + q_2q_3 + q_3q_1 = (2\cdot3 + 3\cdot4 + 4\cdot2)\times10^{-10} = 26\times10^{-10}$.
Step 5: Evaluate the prefactor.
$\dfrac{k}{r} = \dfrac{9\times10^{9}}{0.1} = 9\times10^{10}$.
Step 6: Multiply.
$W = 9\times10^{10} \times 26\times10^{-10} = 9 \times 26 = 234\,\text{J}$.
\[ \boxed{234\ \text{J}} \]