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2.4 - Electric Potential Energy

What It Is

Electric potential energy (UEU_E) is the work an external force has to do to bring charges from infinity to their current positions, moving them at constant velocity. A few things about it: - It’s stored energy that comes from how the charges are arranged. - It can be positive or negative, unlike gravitational PE. - It’s a scalar, not a vector.

The Formula

For two point charges separated by a distance rr:
UE=14πϵ0q1q2r=kq1q2rU_E = \frac{1}{4\pi\epsilon_0}\frac{q_1q_2}{r} = k\frac{q_1q_2}{r}
where: - UEU_E = electric potential energy (J) - k=8.99×109k = 8.99 \times 10^9 N·m²/C² - q1,q2q_1, q_2 = charges (C), and you keep the signs here - rr = separation distance (m)
The reference point is UE=0U_E = 0 when the charges are infinitely far apart.

The Sign of Potential Energy

Unlike Coulomb’s law, do not use absolute values here. The signs matter.
  • UE>0U_E > 0 (positive) happens with like charges (++ or −−):
    • The system has stored energy.
    • The charges repel, so they’d fly apart on their own if released.
  • UE<0U_E < 0 (negative) happens with unlike charges (+− or −+):
    • The system has negative energy, which is a bound state.
    • The charges attract, so they’d come together on their own if released.
Physically: - Positive UEU_E means you have to do work to bring the charges together. - Negative UEU_E means the field does the work as the charges come together.

Force vs. Energy

Property
Coulomb’s Law
Potential Energy
Quantity
Force (vector)
Energy (scalar)
Equation
F=k∥q1q2∥r2F = k\frac{\|q_1q_2\|}{r^2}
UE=kq1q2rU_E = k\frac{q_1q_2}{r}
Distance
∝1/r2\propto 1/r^2
∝1/r\propto 1/r
Signs
Use absolute value
Keep signs

Systems with Multiple Charges

The total electric potential energy is the sum of the PE for each pair of charges. For nn charges, you find the PE of every unique pair:
Utotal=U12+U13+U23+...U_{total} = U_{12} + U_{13} + U_{23} + ...
The number of pairs for nn charges is n(n−1)2\frac{n(n-1)}{2}. So: - 3 charges give 3 pairs (1-2, 1-3, 2-3). - 4 charges give 6 pairs (1-2, 1-3, 1-4, 2-3, 2-4, 3-4).
Count each pair once, and don’t double up.
Why it’s a sum over pairs: remember that UEU_E is the work an outside force does to bring the charges in from infinitely far away. Build the system one charge at a time: - Bring in q1q_1: nothing else is there yet, so it takes no work. - Bring in q2q_2: you’re only working against q1q_1, so W=U12W = U_{12}. - Bring in q3q_3: now you’re working against both, so W=U13+U23W = U_{13} + U_{23}. - Add it up: Utotal=U12+U13+U23U_{total} = U_{12} + U_{13} + U_{23}, exactly one term for each pair.
The order you bring them in doesn’t matter. You always end up with one term per pair. On the AP exam you’ll only have to do this for four or fewer point charges.

Work-Energy Relationship

The work an external force does to move charges is:
Wexternal=ΔUE=Ufinal−UinitialW_{external} = \Delta U_E = U_{final} - U_{initial}
If there’s no external work, energy is conserved:
ΔK+ΔUE=0\Delta K + \Delta U_E = 0
And with external work:
Wexternal=ΔK+ΔUEW_{external} = \Delta K + \Delta U_E
It’s a lot like mechanical energy conservation, just with electric PE instead of gravitational PE.

Practice Problem 1: Two Charges

Problem: Two point charges, q1=+3.0×10−6q_1 = +3.0 \times 10^{-6} C and q2=−5.0×10−6q_2 = -5.0 \times 10^{-6} C, are 0.40 m apart. Find the electric potential energy of the system.
Solution:
UE=kq1q2r=(8.99×109)(+3.0×10−6)(−5.0×10−6)0.40U_E = k\frac{q_1q_2}{r} = (8.99 \times 10^9)\frac{(+3.0 \times 10^{-6})(-5.0 \times 10^{-6})}{0.40}UE=(8.99×109)−15.0×10−120.40U_E = (8.99 \times 10^9)\frac{-15.0 \times 10^{-12}}{0.40}UE=−0.337 JU_E = -0.337 \text{ J}
Answer: −0.34-0.34 J, negative because opposite charges attract.

Practice Problem 2: Three Charges

Problem: Three charges form an equilateral triangle with sides of 0.10 m: - q1=+2.0q_1 = +2.0 μC at vertex A - q2=+2.0q_2 = +2.0 μC at vertex B - q3=−4.0q_3 = -4.0 μC at vertex C
Find the total electric potential energy.
Solution:
Work out the PE for each of the 3 pairs.
Pair 1-2 (both positive, 0.10 m apart):
U12=kq1q2r=(8.99×109)(2.0×10−6)(2.0×10−6)0.10=0.360 JU_{12} = k\frac{q_1q_2}{r} = (8.99 \times 10^9)\frac{(2.0 \times 10^{-6})(2.0 \times 10^{-6})}{0.10} = 0.360 \text{ J}
Pair 1-3 (opposite signs, 0.10 m apart):
U13=kq1q3r=(8.99×109)(2.0×10−6)(−4.0×10−6)0.10=−0.719 JU_{13} = k\frac{q_1q_3}{r} = (8.99 \times 10^9)\frac{(2.0 \times 10^{-6})(-4.0 \times 10^{-6})}{0.10} = -0.719 \text{ J}
Pair 2-3 (opposite signs, 0.10 m apart):
U23=kq2q3r=(8.99×109)(2.0×10−6)(−4.0×10−6)0.10=−0.719 JU_{23} = k\frac{q_2q_3}{r} = (8.99 \times 10^9)\frac{(2.0 \times 10^{-6})(-4.0 \times 10^{-6})}{0.10} = -0.719 \text{ J}
Total:
Utotal=U12+U13+U23=0.360−0.719−0.719=−1.08 JU_{total} = U_{12} + U_{13} + U_{23} = 0.360 - 0.719 - 0.719 = -1.08 \text{ J}
Answer: −1.1-1.1 J, negative overall because the attractive interactions win out.

Practice Problem 3: Work and Energy

Problem: An electron is released from rest 2.0 mm from a fixed proton. What is the electron’s speed when it’s 1.0 mm from the proton?
Given: me=9.11×10−31m_e = 9.11 \times 10^{-31} kg.
Solution:
Energy conservation with no external forces:
Ki+Ui=Kf+UfK_i + U_i = K_f + U_f
Initial state (at rest, ri=2.0×10−3r_i = 2.0 \times 10^{-3} m):
Ki=0K_i = 0Ui=kqeqpri=(8.99×109)(−e)(+e)2.0×10−3U_i = k\frac{q_e q_p}{r_i} = (8.99 \times 10^9)\frac{(-e)(+e)}{2.0 \times 10^{-3}}Ui=−(8.99×109)(1.60×10−19)22.0×10−3=−1.15×10−25 JU_i = -(8.99 \times 10^9)\frac{(1.60 \times 10^{-19})^2}{2.0 \times 10^{-3}} = -1.15 \times 10^{-25} \text{ J}
Final state (rf=1.0×10−3r_f = 1.0 \times 10^{-3} m):
Kf=12mev2K_f = \frac{1}{2}m_e v^2Uf=kqeqprf=(8.99×109)(−e)(+e)1.0×10−3U_f = k\frac{q_e q_p}{r_f} = (8.99 \times 10^9)\frac{(-e)(+e)}{1.0 \times 10^{-3}}Uf=−(8.99×109)(1.60×10−19)21.0×10−3=−2.30×10−25 JU_f = -(8.99 \times 10^9)\frac{(1.60 \times 10^{-19})^2}{1.0 \times 10^{-3}} = -2.30 \times 10^{-25} \text{ J}
Putting it together:
0+(−1.15×10−25)=12mev2+(−2.30×10−25)0 + (-1.15 \times 10^{-25}) = \frac{1}{2}m_e v^2 + (-2.30 \times 10^{-25})12mev2=−1.15×10−25−(−2.30×10−25)=1.15×10−25 J\frac{1}{2}m_e v^2 = -1.15 \times 10^{-25} - (-2.30 \times 10^{-25}) = 1.15 \times 10^{-25} \text{ J}v=2(1.15×10−25)9.11×10−31=2.52×105=502 m/sv = \sqrt{\frac{2(1.15 \times 10^{-25})}{9.11 \times 10^{-31}}} = \sqrt{2.52 \times 10^5} = 502 \text{ m/s}
Answer: 502 m/s.
The physical read on this: the PE dropped (got more negative), so the KE went up. The electron accelerates toward the proton.

Practice Problem 4: Work to Bring In a Charge

Problem: Two +2.0+2.0 μC charges are fixed 0.60 m apart. An outside force brings a third charge from very far away to the point halfway between them, moving it at constant speed. How much work does the outside force do if the third charge is: (a) +1.0+1.0 μC? (b) −1.0-1.0 μC?
Solution:
The charge moves at constant speed, so ΔK=0\Delta K = 0 and the outside work equals the change in the system’s potential energy. Only the two new pairs are added (U12U_{12} doesn’t change), and the midpoint is 0.30 m from each fixed charge:
W=ΔUE=U13+U23=2kq1q3rW = \Delta U_E = U_{13} + U_{23} = 2k\frac{q_1q_3}{r}
  1. For q3=+1.0q_3 = +1.0 μC:
      2. W=2(8.99×109)(2.0×10−6)(1.0×10−6)0.30=+0.12 JW = 2(8.99 \times 10^9)\frac{(2.0 \times 10^{-6})(1.0 \times 10^{-6})}{0.30} = +0.12 \text{ J}
  1. For q3=−1.0q_3 = -1.0 μC:
      2. W=2(8.99×109)(2.0×10−6)(−1.0×10−6)0.30=−0.12 JW = 2(8.99 \times 10^9)\frac{(2.0 \times 10^{-6})(-1.0 \times 10^{-6})}{0.30} = -0.12 \text{ J}
Answer: (a) +0.12+0.12 J, (b) −0.12-0.12 J.
The read on this: in (a) the fixed charges repel the incoming charge, so you have to push it in, and the system’s PE goes up from U12=0.060U_{12} = 0.060 J to 0.18 J. In (b) they attract it, so you have to hold it back the whole way in (negative work), and the system’s PE goes down.

Energy Graphs

2.4 - Electric Potential Energy
Key features to notice: - Both curves head toward 0 as r→∞r \to \infty, which is the reference point. - Like charges always have U>0U > 0. - Unlike charges always have U<0U < 0. - Both curves get steeper at small rr, where the interaction is stronger.

Common Mistakes

  • Using absolute values. Unlike force, PE keeps its sign. Use UE=kq1q2rU_E = k\frac{q_1q_2}{r} with the real signs, not absolute values.
  • Forgetting the reference point. Zero PE is at infinite separation, not at some finite distance.
  • Double counting pairs. For 3 charges, there are 3 pairs, not 6. Count each pair once.
  • Confusing PE and force. Force goes as 1/r21/r^2 (inverse square), while PE goes as 1/r1/r (just inverse, not squared).
  • Misreading the sign. Negative PE means the system has less energy than when the charges are infinitely far apart. To pull them apart, an outside force has to add energy (do positive work). That’s what makes it a bound, attractive state.
  • Thinking negative PE is impossible. Negative values are perfectly normal and physical for attractive setups.
  • Mixing up initial and final. W=ΔUE=Uf−UiW = \Delta U_E = U_f - U_i, so check which is which.
  • Not using energy conservation. When charges move freely with no external force, use Ki+Ui=Kf+UfK_i + U_i = K_f + U_f.
  • Forgetting that kinetic energy changes too. If a charge moves, both PE and KE change, so use energy conservation to link them.

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