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2.3 - Conservation of Electric Energy

Electric Potential Energy and Work

When a charged object moves between two spots with different electric potentials, the change in electric potential energy is:
ΔUE=qΔV\Delta U_E = q\Delta V
where: - ΔUE\Delta U_E = change in electric potential energy (J). - qq = charge of the object (C). - ΔV=Vf−Vi\Delta V = V_f - V_i = potential difference (V).
The potential energy belongs to the object-field system: the charged object together with the field (really, the object plus the charges that make the field). If you pick just the object as your system instead, it has no potential energy, and the field is an outside force doing work W=−qΔVW = -q\Delta V on it. Both views give the same change in kinetic energy.
How it stacks up against gravitational PE:
Gravitational PE is Ug=mghU_g = mgh, where you move a mass through a height difference. Electric PE works the same way, except you’re moving a charge through a potential difference.
Gravitational
Electric
Ug=mghU_g = mgh
UE=qVU_E = qV
Move mass through a height
Move charge through a potential
ΔUg=mgΔh\Delta U_g = mg\Delta h
ΔUE=qΔV\Delta U_E = q\Delta V

Conservation of Energy with Electric Potential

Conservation of energy stretches to cover electric potential energy too:
ΔEtotal=0\Delta E_{total} = 0
Written out in full:
ΔK+ΔUg+ΔUs+ΔUE=0\Delta K + \Delta U_g + \Delta U_s + \Delta U_E = 0
where: - ΔK\Delta K = change in kinetic energy. - ΔUg\Delta U_g = change in gravitational potential energy. - ΔUs\Delta U_s = change in spring or elastic potential energy. - ΔUE\Delta U_E = change in electric potential energy.
Rearranged for actually solving problems:
Ki+Ug,i+Us,i+UE,i=Kf+Ug,f+Us,f+UE,fK_i + U_{g,i} + U_{s,i} + U_{E,i} = K_f + U_{g,f} + U_{s,f} + U_{E,f}
Or, when only electric and kinetic energy are in play:
Ki+UE,i=Kf+UE,fK_i + U_{E,i} = K_f + U_{E,f}

Work and Energy in Electric Fields

The sign of the work depends on whether the charge is moving the way it naturally wants to, or getting shoved against its tendency.

Positive Work (the Field Does the Work)

When charges move the way they “want” to: - A positive charge heads toward lower potential (with the field). - A negative charge heads toward higher potential (against the field). - The electric field does positive work. - Electric PE drops, kinetic energy climbs.
2.3 - Conservation of Electric Energy

Negative Work (an External Force Does the Work)

When something outside pushes charges against their natural tendency: - A positive charge gets forced toward higher potential (against the field). - A negative charge gets forced toward lower potential (with the field). - The external force does positive work, and the field does negative work. - Electric PE goes up, and kinetic energy goes down if it started from rest.
2.3 - Conservation of Electric Energy
The key relationships:
Wfield=−ΔUE=−qΔVW_{field} = -\Delta U_E = -q\Delta VWexternal=ΔUE=qΔVW_{external} = \Delta U_E = q\Delta V

Example Problem 1: Proton Accelerated Through a Potential Difference

Problem: A proton (mass mp=1.67×10−27m_p = 1.67 \times 10^{-27} kg) accelerates from rest through a potential difference of 4.5×1064.5 \times 10^6 V.
  1. How much kinetic energy did the proton pick up?
  1. If it started at rest, how fast is it going?
2.3 - Conservation of Electric Energy

Solution

(a) Kinetic Energy Gained
Use the link between potential difference and energy change:
ΔUE=qΔV\Delta U_E = q\Delta V
A proton has charge q=+1.6×10−19q = +1.6 \times 10^{-19} C. A positive charge speeds up by moving to lower potential, so the proton moves through ΔV=−4.5×106\Delta V = -4.5 \times 10^6 V.
ΔUE=(1.6×10−19)(−4.5×106)\Delta U_E = (1.6 \times 10^{-19})(-4.5 \times 10^6)ΔUE=−7.2×10−13 J\Delta U_E = -7.2 \times 10^{-13} \text{ J}
Since it starts from rest and the only thing happening is electric PE turning into kinetic energy:
ΔK=−ΔUE\Delta K = -\Delta U_EKf−0=−ΔUEK_f - 0 = -\Delta U_E
So the kinetic energy gained is:
Kf=7.2×10−13 JK_f = 7.2 \times 10^{-13} \text{ J}
Answer (a): the proton picked up 7.2×10−137.2 \times 10^{-13} J of kinetic energy.
(b) Final Speed
From the kinetic energy formula:
K=12mv2K = \frac{1}{2}mv^2
Solve for vv:
v=2Kmv = \sqrt{\frac{2K}{m}}v=2(7.2×10−13)1.67×10−27v = \sqrt{\frac{2(7.2 \times 10^{-13})}{1.67 \times 10^{-27}}}v=8.62×1014v = \sqrt{8.62 \times 10^{14}}v=2.94×107 m/sv = 2.94 \times 10^7 \text{ m/s}
Answer (b): the proton ends up moving at roughly 2.9×1072.9 \times 10^7 m/s.
For context: the speed of light is c=3×108c = 3 \times 10^8 m/s, so this proton is cruising at about 10% of light speed.
Shortcut with electron volts: 1 eV =1.60×10−19= 1.60 \times 10^{-19} J is the energy a particle with charge ee gains moving through 1 V. So a proton accelerated through 4.5×1064.5 \times 10^6 V gains 4.5×1064.5 \times 10^6 eV (4.5 MeV) of kinetic energy, and (4.5×106)(1.60×10−19 J)=7.2×10−13(4.5 \times 10^6)(1.60 \times 10^{-19} \text{ J}) = 7.2 \times 10^{-13} J, the same answer as part (a).

Example Problem 2: Electron in a Uniform Field

Problem: An electron (mass me=9.11×10−31m_e = 9.11 \times 10^{-31} kg, charge q=−1.6×10−19q = -1.6 \times 10^{-19} C) is released from rest in a uniform electric field of magnitude E=2.0×104E = 2.0 \times 10^4 N/C. After traveling 0.50 m in the direction opposite to the field, how fast is it moving?
2.3 - Conservation of Electric Energy

Solution

Approach: energy conservation
The electron moves opposite to the field, which means it’s heading from lower potential to higher potential. For a negative charge, that’s a drop in potential energy.
The potential difference it travels through:
ΔV=Ed=(2.0×104)(0.50)=1.0×104 V\Delta V = Ed = (2.0 \times 10^4)(0.50) = 1.0 \times 10^4 \text{ V}
The change in electric potential energy:
ΔUE=qΔV=(−1.6×10−19)(1.0×104)\Delta U_E = q\Delta V = (-1.6 \times 10^{-19})(1.0 \times 10^4)ΔUE=−1.6×10−15 J\Delta U_E = -1.6 \times 10^{-15} \text{ J}
That negative value means the PE went down, and that energy shows up as kinetic energy.
Using energy conservation from rest:
Kf=−ΔUE=1.6×10−15 JK_f = -\Delta U_E = 1.6 \times 10^{-15} \text{ J}
Solve for the speed:
v=2Kfme=2(1.6×10−15)9.11×10−31v = \sqrt{\frac{2K_f}{m_e}} = \sqrt{\frac{2(1.6 \times 10^{-15})}{9.11 \times 10^{-31}}}v=3.51×1015v = \sqrt{3.51 \times 10^{15}}v=5.93×107 m/sv = 5.93 \times 10^7 \text{ m/s}
Answer: the electron is moving at about 5.9×1075.9 \times 10^7 m/s, roughly 20% of light speed.

Example Problem 3: Proton in a Uniform Field

Problem: A proton (mass mp=1.67×10−27m_p = 1.67 \times 10^{-27} kg, charge q=+1.6×10−19q = +1.6 \times 10^{-19} C) sits in a uniform electric field of magnitude E=5.0×105E = 5.0 \times 10^5 N/C pointing upward, then gets released from rest.
  1. After it has moved 0.20 m upward, how fast is it going?
  1. What potential difference did it move through?
2.3 - Conservation of Electric Energy

Solution

(a) Final Speed
The proton is positive, so it accelerates upward, right along the field.
Using energy:
The work done by the field:
W=qEd=(1.6×10−19)(5.0×105)(0.20)W = qEd = (1.6 \times 10^{-19})(5.0 \times 10^5)(0.20)W=1.6×10−14 JW = 1.6 \times 10^{-14} \text{ J}
That work turns into kinetic energy:
Kf=W=1.6×10−14 JK_f = W = 1.6 \times 10^{-14} \text{ J}12mpv2=1.6×10−14\frac{1}{2}m_pv^2 = 1.6 \times 10^{-14}v=2(1.6×10−14)1.67×10−27v = \sqrt{\frac{2(1.6 \times 10^{-14})}{1.67 \times 10^{-27}}}v=1.916×1013v = \sqrt{1.916 \times 10^{13}}v=4.38×106 m/sv = 4.38 \times 10^6 \text{ m/s}
Answer (a): v=4.4×106v = 4.4 \times 10^6 m/s.
(b) Potential Difference
For a uniform field, moving along the field direction:
ΔV=−Ed=−(5.0×105)(0.20)\Delta V = -Ed = -(5.0 \times 10^5)(0.20)ΔV=−1.0×105 V\Delta V = -1.0 \times 10^5 \text{ V}
Answer (b): the proton moved through a potential difference of −100-100 kV, so the potential dropped by 100 kV.

Example Problem 4: Closest Approach

Problem: A small sphere (m=2.0m = 2.0 g, q=+3.0q = +3.0 μC) slides on a level, frictionless surface straight toward a fixed charge Q=+5.0Q = +5.0 μC. When it’s 0.90 m away, its speed is 15 m/s.
  1. How close does the sphere get to the fixed charge?
  1. How fast is it moving once it has been pushed back out to very far away?
2.3 - Conservation of Electric Energy

Solution

The field of a point charge isn’t uniform, so ΔV=−Ed\Delta V = -Ed won’t work here. Use energy conservation with UE=kqQrU_E = k\frac{qQ}{r}. The surface is level and frictionless, so only the electric force does work.
(a) Closest Approach
At the closest point the sphere is momentarily at rest, so all of its energy is electric potential energy:
12mvi2+kqQri=0+kqQrmin\frac{1}{2}mv_i^2 + k\frac{qQ}{r_i} = 0 + k\frac{qQ}{r_{min}}
First, kqQ=(9.0×109)(3.0×10−6)(5.0×10−6)=0.135kqQ = (9.0 \times 10^9)(3.0 \times 10^{-6})(5.0 \times 10^{-6}) = 0.135 J·m. The total energy is:
12(2.0×10−3)(15)2+0.1350.90=0.225 J+0.15 J=0.375 J\frac{1}{2}(2.0 \times 10^{-3})(15)^2 + \frac{0.135}{0.90} = 0.225 \text{ J} + 0.15 \text{ J} = 0.375 \text{ J}rmin=0.135 J⋅m0.375 J=0.36 mr_{min} = \frac{0.135 \text{ J}\cdot\text{m}}{0.375 \text{ J}} = 0.36 \text{ m}
Answer (a): the sphere gets within 0.36 m of the fixed charge, then gets pushed back.
(b) Speed Far Away
Very far away, UE→0U_E \to 0, so all 0.375 J is kinetic energy:
v=2Km=2(0.375)2.0×10−3=19 m/sv = \sqrt{\frac{2K}{m}} = \sqrt{\frac{2(0.375)}{2.0 \times 10^{-3}}} = 19 \text{ m/s}
Answer (b): about 19 m/s. It ends up faster than it was at 0.90 m because the 0.15 J of potential energy it had there also turns into kinetic energy.

Speed Comparisons

It’s handy to size these particle speeds up against the speed of light, since that’s what tells you when relativity starts to matter:
c=3.0×108 m/sc = 3.0 \times 10^8 \text{ m/s}
From our examples: - Proton at 2.9×1072.9 \times 10^7 m/s: v/c=0.097v/c = 0.097, so about 10% of light speed. - Electron at 5.9×1075.9 \times 10^7 m/s: v/c=0.197v/c = 0.197, so about 20% of light speed. - Proton at 4.4×1064.4 \times 10^6 m/s: v/c=0.015v/c = 0.015, so about 1.5% of light speed.
Rule of thumb: classical mechanics (K=12mv2K = \frac{1}{2}mv^2) works well as long as vv is a small fraction of cc. For a given kinetic energy, it overestimates the speed by about 0.4% at 10% of light speed and about 1.5% at 20% (the electron above), so these answers are close enough. The error grows quickly after that (about 11% at half the speed of light), and you’d need special relativity, which isn’t part of this course.

How Speed Scales with the Potential Difference

For a particle that starts from rest, 12mv2=∣qΔV∣\frac{1}{2}mv^2 = |q\Delta V|, so v=2∣qΔV∣mv = \sqrt{\frac{2|q\Delta V|}{m}} and v∝∣ΔV∣v \propto \sqrt{|\Delta V|}. Four times the potential difference gives twice the speed, not four times.
Example: An electron accelerated from rest through 100 V reaches:
v=2(1.60×10−19 C)(100 V)9.11×10−31 kg=5.9×106 m/sv = \sqrt{\frac{2(1.60 \times 10^{-19} \text{ C})(100 \text{ V})}{9.11 \times 10^{-31} \text{ kg}}} = 5.9 \times 10^6 \text{ m/s}
Through 400 V it reaches 4=2\sqrt{4} = 2 times that, 1.2×1071.2 \times 10^7 m/s.

Energy Bar Charts

Drawing out the energy helps make the transformations click:
2.3 - Conservation of Electric Energy

Problem-Solving Strategy

For conservation of electric energy problems:
  1. Pin down the initial and final states. Where does the particle start and where does it end up?
  1. Figure out what you’ve been given. Do you have ΔV\Delta V, or EE and dd, or both?
  1. Pick a zero reference. Usually infinity, or one of the spots in the problem.
  1. Apply energy conservation:
      2. Ki+UE,i=Kf+UE,fK_i + U_{E,i} = K_f + U_{E,f}
  1. Solve for the unknown, which is usually speed, distance, or potential difference.
  1. Check the signs. Did the PE go up or down? Does that make sense physically?

Key Formulas Summary

Relationship
Formula
PE change
ΔUE=qΔV\Delta U_E = q\Delta V
Work by the field
Wfield=−qΔV=−ΔUEW_{field} = -q\Delta V = -\Delta U_E
Uniform field
ΔV=−Ed\Delta V = -Ed (moving with the field)
Energy conservation
Ki+UE,i=Kf+UE,fK_i + U_{E,i} = K_f + U_{E,f}
Starting from rest
Kf=−ΔUE=−qΔVK_f = -\Delta U_E = -q\Delta V
Speed from energy
v=2Kmv = \sqrt{\frac{2K}{m}}

Sign Conventions

For the potential difference ΔV=Vf−Vi\Delta V = V_f - V_i:
Particle
Motion Direction
ΔV\Delta V
ΔUE\Delta U_E
Speed Change
Positive
With the field (toward −)
Negative
Negative
Speeds up
Positive
Against the field (toward +)
Positive
Positive
Slows down
Negative
Against the field (toward +)
Positive
Negative
Speeds up
Negative
With the field (toward −)
Negative
Positive
Slows down
The rule: particles speed up whenever ΔUE<0\Delta U_E < 0, meaning their PE drops.

Common Mistakes

  1. Forgetting the sign of the charge in ΔUE=qΔV\Delta U_E = q\Delta V. That qq carries its sign. For an electron it’s q=−1.6×10−19q = -1.6 \times 10^{-19} C, not positive. The sign flows straight through to ΔUE\Delta U_E and determines whether energy is gained or lost.
  1. Misreading “accelerated through X volts.” When a problem says a particle is “accelerated through 500 V,” it means the particle moves through a potential difference that speeds it up. For a proton that means going to lower potential, and for an electron it means going to higher potential. The magnitude is 500 V, but the sign of ΔV\Delta V depends on which particle you’ve got.
  1. Mixing up work by the field with work by an external force. The field does positive work when particles move naturally (W=−ΔUEW = -\Delta U_E). An external force does positive work when it drags particles against their tendency (W=+ΔUEW = +\Delta U_E). Opposite signs.
  1. Misapplying conservation of energy. ΔK=−ΔUE\Delta K = -\Delta U_E only holds when nothing else is doing work. If gravity is in the picture, you have to include ΔUg\Delta U_g too.
  1. Forgetting the square root when finding speed. Once you’ve got K=12mv2K = \frac{1}{2}mv^2, you still need to take the square root: v=2Kmv = \sqrt{\frac{2K}{m}}, not v=2Kmv = \frac{2K}{m}.

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