Question

Three charges q, q and Q ($q=+20~\mu C$ and $Q=+10~\mu C$) are placed on the circumference of a circle of radius $10\sqrt3$ cm. If the distance between any two charges is same, then the total electrostatic potential energy of the system of the three charges is

• A. 48 J
• B. 36 J
• C. 24 J
• D. 12 J

Hint:

For a system of charges, add the potential energies of all distinct pairs only once. Do not count any pair twice.

Answer 1

The Correct Option is C

Concept: The electrostatic potential energy of a system of charges is \[ U=\sum \frac{kq_iq_j}{r_{ij}} \] where the summation is taken over all unique pairs of charges.

Step 1: Determine the distance between charges.
The three charges are equally separated on the circumference. Hence they form an equilateral triangle. For an equilateral triangle, \[ a=\sqrt3R \] Given, \[ R=10\sqrt3\,cm \] Thus, \[ a=\sqrt3(10\sqrt3) \] \[ a=30\,cm \] \[ a=0.3\,m \]

Step 2: Write the total potential energy.
\[ U= \frac{kq^2}{a} +\frac{kqQ}{a} +\frac{kqQ}{a} \] \[ U= \frac{k}{a} (q^2+2qQ) \]

Step 3: Substitute the numerical values.
\[ q=20\times10^{-6}C \] \[ Q=10\times10^{-6}C \] \[ U= \frac{9\times10^9}{0.3} \left[ (20\times10^{-6})^2 + 2(20\times10^{-6})(10\times10^{-6}) \right] \] \[ = 3\times10^{10} \times 8\times10^{-10} \] \[ U=24J \] \[ \boxed{24J} \]