Electrostatics: Step-by-Step Worked Examples (12)

hard 3 min read

Question

Three point charges +2μC+2 \, \mu C, 3μC-3 \, \mu C, and +4μC+4 \, \mu C are placed at the corners of an equilateral triangle of side 0.1m0.1 \, \text{m}. Find the net electrostatic potential energy of the system.

Solution — Step by Step

For each pair of charges, the potential energy is:

Uij=14πϵ0qiqjrijU_{ij} = \frac{1}{4\pi\epsilon_0} \cdot \frac{q_i q_j}{r_{ij}}

For three charges, total PE is the sum of all three pairs: U=U12+U13+U23U = U_{12} + U_{13} + U_{23}.

All three pairs have the same separation r=0.1mr = 0.1 \, \text{m}. The charge products:

  • Pair 1-2: (+2)(3)=6μC2(+2)(-3) = -6 \, \mu C^2
  • Pair 1-3: (+2)(+4)=+8μC2(+2)(+4) = +8 \, \mu C^2
  • Pair 2-3: (3)(+4)=12μC2(-3)(+4) = -12 \, \mu C^2

Sum: 6+812=10μC2=10×1012C2-6 + 8 - 12 = -10 \, \mu C^2 = -10 \times 10^{-12} \, C^2.

k=1/(4πϵ0)=9×109N m2/C2k = 1/(4\pi\epsilon_0) = 9 \times 10^9 \, \text{N m}^2/\text{C}^2.

U=kqiqjr=9×109×(10×1012)0.1U = \frac{k \cdot \sum q_i q_j}{r} = \frac{9 \times 10^9 \times (-10 \times 10^{-12})}{0.1}

U=9×1020.1=0.9JU = \frac{-9 \times 10^{-2}}{0.1} = -0.9 \, \text{J}

Total electrostatic PE: U=0.9JU = -0.9 \, \text{J}.

The negative sign means the system is bound — we’d need to do positive work to separate all three charges to infinity.

Why This Works

Potential energy is a scalar, so we just add the contributions. No vector book-keeping, no angle calculations — that is what makes PE problems faster than force or field problems.

Each pair contributes independently. The factor of 1/21/2 that some students remember from U=qiVi/2U = \sum q_i V_i / 2 does NOT appear when we sum pairwise — because the sum of pairs already counts each bond once.

Speed shortcut: When all separations are equal (equilateral triangle, square with diagonal asked separately, etc.), factor out k/rk/r and just sum the charge products. We get the answer in three lines.

Alternative Method — Using Potentials

We could compute the potential at each corner due to the other two charges, multiply by that corner’s charge, and divide by 2:

U=12iqiViU = \frac{1}{2} \sum_i q_i V_i

This works but doubles the arithmetic. Use it only when the question gives potentials directly.

Common Mistake

Students often forget the signs of charges and treat all products as positive — getting +0.9J+0.9 \, \text{J} instead of 0.9J-0.9 \, \text{J}. The sign is half the answer.

Another classic: counting each pair twice. If we sum U12+U21+U13+U31+U23+U32U_{12} + U_{21} + U_{13} + U_{31} + U_{23} + U_{32}, we get double the right answer. Always count each pair once.

JEE Advanced 2022 had a four-charge configuration (square corners). Six pairs to count, with two pairs at separation aa and four at a2a\sqrt{2}. Same template, just more bookkeeping. Top scorers solved it in under 90 seconds by tabulating cleanly.

Want to master this topic?

Read the complete guide with more examples and exam tips.

Go to full topic guide →

Try These Next