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2.7 - Conservation of Electric Energy

Change in Electric Potential Energy

When a charge moves through a potential difference, its electric potential energy changes:
ΔUE=qΔV\Delta U_E = q\Delta V
where: - ΔUE\Delta U_E = change in electric potential energy (J) - qq = charge (C), and you keep the sign - ΔV=Vf−Vi\Delta V = V_f - V_i = potential difference (V)
The sign matters: - qΔV>0q\Delta V > 0 means the PE increases, so the charge gains PE. - qΔV<0q\Delta V < 0 means the PE decreases, so the charge loses PE.
Whose energy is it? Potential energy belongs to a system: the charged object plus the field it’s moving through (the CED calls this the object-field system). If you pick the charged object alone as your system, it has no potential energy. Instead, the field is outside the system and does work on it, W=−qΔVW = -q\Delta V. Both views give the same change in kinetic energy.

Conservation of Energy

Energy is conserved when charges move through electric fields:
ΔK+ΔUE=Wexternal\Delta K + \Delta U_E = W_{external}
For an isolated system, where there’s no external work and only electric forces act:
ΔK+ΔUE=0\Delta K + \Delta U_E = 0Kf−Ki+qΔV=0K_f - K_i + q\Delta V = 0Kf=Ki−qΔVK_f = K_i - q\Delta V
What it means: - If PE decreases, KE increases, so the charge speeds up. - If PE increases, KE decreases, so the charge slows down.

Common Scenarios

1. A Charge Released from Rest

Starting: Ki=0K_i = 0, at potential ViV_i. Ending: Kf=?K_f = ?, at potential VfV_f.
Kf=−q(Vf−Vi)=q(Vi−Vf)K_f = -q(V_f - V_i) = q(V_i - V_f)
For speed:
12mv2=q(Vi−Vf)\frac{1}{2}mv^2 = q(V_i - V_f)v=2q(Vi−Vf)mv = \sqrt{\frac{2q(V_i - V_f)}{m}}

2. Positive Charge Movement

Moving high to low potential (Vi>VfV_i > V_f, so ΔV<0\Delta V < 0): - ΔUE=qΔV<0\Delta U_E = q\Delta V < 0, so PE decreases. - ΔK>0\Delta K > 0, so KE increases. - It speeds up on its own.
Moving low to high potential (Vi<VfV_i < V_f, so ΔV>0\Delta V > 0): - ΔUE=qΔV>0\Delta U_E = q\Delta V > 0, so PE increases. - ΔK<0\Delta K < 0, so KE decreases. - It slows down, or it needs external work.

3. Negative Charge Movement

Moving high to low potential (ΔV<0\Delta V < 0): - ΔUE=qΔV>0\Delta U_E = q\Delta V > 0, since negative times negative is positive, so PE increases. - ΔK<0\Delta K < 0, so it slows down. - It needs external work to push against the field.
Moving low to high potential (ΔV>0\Delta V > 0): - ΔUE=qΔV<0\Delta U_E = q\Delta V < 0, since negative times positive is negative, so PE decreases. - ΔK>0\Delta K > 0, so it speeds up. - It accelerates naturally toward high potential.

Energy in Capacitors

For a charge moving between capacitor plates, the field is uniform: E=ΔVdE = \frac{\Delta V}{d}.
For a charge qq moving a distance dd across the full potential difference:
ΔUE=qΔV\Delta U_E = q\Delta V
Using kinematics with constant acceleration:
a=qEma = \frac{qE}{m}
Both routes give the same answer, but energy is usually faster. For example, a proton is released from rest at the positive plate of a capacitor with 200 V across a 4.0 mm gap. How fast is it going when it reaches the negative plate?
Energy route (the proton moves from high to low VV, so ΔV=−200\Delta V = -200 V):
12mv2=−qΔV⇒v=2(1.60×10−19)(200)1.67×10−27=1.96×105 m/s\frac{1}{2}mv^2 = -q\Delta V \quad\Rightarrow\quad v = \sqrt{\frac{2(1.60 \times 10^{-19})(200)}{1.67 \times 10^{-27}}} = 1.96 \times 10^5 \text{ m/s}
Kinematics route:
E=ΔVd=2000.0040=5.0×104 N/C,a=qEm=(1.60×10−19)(5.0×104)1.67×10−27=4.79×1012 m/s2E = \frac{\Delta V}{d} = \frac{200}{0.0040} = 5.0 \times 10^4 \text{ N/C}, \quad a = \frac{qE}{m} = \frac{(1.60 \times 10^{-19})(5.0 \times 10^4)}{1.67 \times 10^{-27}} = 4.79 \times 10^{12} \text{ m/s}^2v=2ad=2(4.79×1012)(0.0040)=1.96×105 m/sv = \sqrt{2ad} = \sqrt{2(4.79 \times 10^{12})(0.0040)} = 1.96 \times 10^5 \text{ m/s}
Same answer either way, about 2.0×1052.0 \times 10^5 m/s. The energy route skips finding EE and aa, and it still works when the field isn’t uniform, where the constant-acceleration equations don’t apply.

Practice Problem 1: Electron Acceleration

Problem: An electron (charge −e-e, mass 9.11×10−319.11 \times 10^{-31} kg) is accelerated from rest through a potential difference of 500 V. Find its final speed.
Solution:
Energy conservation:
Kf+ΔUE=KiK_f + \Delta U_E = K_i
Starting from rest, Ki=0K_i = 0:
12mv2=−qΔV\frac{1}{2}mv^2 = -q\Delta V
For an electron, q=−e=−1.60×10−19q = -e = -1.60 \times 10^{-19} C:
12mv2=−(−e)(500)=500e\frac{1}{2}mv^2 = -(-e)(500) = 500ev=2(500e)m=2(500)(1.60×10−19)9.11×10−31v = \sqrt{\frac{2(500e)}{m}} = \sqrt{\frac{2(500)(1.60 \times 10^{-19})}{9.11 \times 10^{-31}}}v=1.60×10−169.11×10−31=1.76×1014v = \sqrt{\frac{1.60 \times 10^{-16}}{9.11 \times 10^{-31}}} = \sqrt{1.76 \times 10^{14}}v=1.33×107 m/sv = 1.33 \times 10^7 \text{ m/s}
Answer: 1.33×1071.33 \times 10^7 m/s, about 4.4% of the speed of light.
Note: the electron gains KE equal to 500500 eV =500×1.60×10−19= 500 \times 1.60 \times 10^{-19} J.

Practice Problem 2: Proton Slowing Down

Problem: A proton with an initial speed of 3.0×1053.0 \times 10^5 m/s enters a region where it moves from 100 V to 400 V. Find its final speed.
Given: mp=1.67×10−27m_p = 1.67 \times 10^{-27} kg, qp=+eq_p = +e.
Solution:
Energy conservation:
Kf+Uf=Ki+UiK_f + U_f = K_i + U_i12mvf2+qVf=12mvi2+qVi\frac{1}{2}mv_f^2 + qV_f = \frac{1}{2}mv_i^2 + qV_i12mvf2=12mvi2+q(Vi−Vf)\frac{1}{2}mv_f^2 = \frac{1}{2}mv_i^2 + q(V_i - V_f)12mvf2=12(1.67×10−27)(3.0×105)2+(1.60×10−19)(100−400)\frac{1}{2}mv_f^2 = \frac{1}{2}(1.67 \times 10^{-27})(3.0 \times 10^5)^2 + (1.60 \times 10^{-19})(100 - 400)12mvf2=7.52×10−17+(1.60×10−19)(−300)\frac{1}{2}mv_f^2 = 7.52 \times 10^{-17} + (1.60 \times 10^{-19})(-300)12mvf2=7.52×10−17−4.80×10−17=2.72×10−17 J\frac{1}{2}mv_f^2 = 7.52 \times 10^{-17} - 4.80 \times 10^{-17} = 2.72 \times 10^{-17} \text{ J}vf=2(2.72×10−17)1.67×10−27=3.26×1010v_f = \sqrt{\frac{2(2.72 \times 10^{-17})}{1.67 \times 10^{-27}}} = \sqrt{3.26 \times 10^{10}}vf=1.8×105 m/sv_f = 1.8 \times 10^5 \text{ m/s}
Answer: 1.8×1051.8 \times 10^5 m/s.
Check: a positive charge moving to higher potential gains PE and loses KE, so it slows down. ✓

Practice Problem 3: Stopping Potential

Problem: An electron with 2.0 eV of kinetic energy enters a region with a uniform electric field. Through what potential difference does it need to travel to come to rest?
Solution:
Energy conservation, coming to rest so Kf=0K_f = 0:
0=Ki−qΔV0 = K_i - q\Delta VΔV=Kiq\Delta V = \frac{K_i}{q}
For an electron, q=−eq = -e and Ki=2.0K_i = 2.0 eV =2.0e= 2.0e J:
ΔV=2.0e−e=−2.0 V\Delta V = \frac{2.0e}{-e} = -2.0 \text{ V}
Answer: the electron has to travel through a potential difference of −2.0 V, moving to a point 2.0 V lower in potential.
The read on this: an electron, being negative, naturally speeds up when it moves toward higher potential. To stop it, it has to move the other way, toward lower potential, gaining 2.0 eV of PE while losing 2.0 eV of KE.

Practice Problem 4: Multiple Points

Problem: A +3.0 μC charge moves from point A (VA=50V_A = 50 V) to point B (VB=20V_B = 20 V). (a) What is the change in electric potential energy? (b) If the charge started from rest at A, what is its kinetic energy at B?
Solution:
  1. Change in PE:
      2. ΔUE=qΔV=q(VB−VA)\Delta U_E = q\Delta V = q(V_B - V_A)ΔUE=(3.0×10−6)(20−50)=(3.0×10−6)(−30)\Delta U_E = (3.0 \times 10^{-6})(20 - 50) = (3.0 \times 10^{-6})(-30)ΔUE=−9.0×10−5 J=−90 μJ\Delta U_E = -9.0 \times 10^{-5} \text{ J} = -90 \text{ μJ}
Answer: the PE dropped by 90 μJ, with the negative sign meaning a decrease.
  1. Released from rest, Ki=0K_i = 0:
      2. Kf=−ΔUE=−(−9.0×10−5)=9.0×10−5 JK_f = -\Delta U_E = -(-9.0 \times 10^{-5}) = 9.0 \times 10^{-5} \text{ J}
Answer: KB=90K_B = 90 μJ.
The read: a positive charge moving from high to low potential loses PE and gains KE.

Common Mistakes

  • Forgetting the charge sign. For electrons, q=−eq = -e, which is negative. That flips the sign of ΔUE\Delta U_E.
  • Getting the direction of ΔV wrong. ΔV=Vf−Vi\Delta V = V_f - V_i, final minus initial. Stay consistent.
  • Thinking a negative ΔU is impossible. A negative ΔUE\Delta U_E just means the PE decreased, which is totally normal and common.
  • Mixing up positive and negative charge behavior. A positive charge accelerates high to low V, while a negative charge accelerates low to high V.
  • Forgetting conservation of energy. Use Ki+Ui=Kf+UfK_i + U_i = K_f + U_f, or the equivalent ΔK=−ΔUE\Delta K = -\Delta U_E.
  • Using the wrong reference. Make sure ViV_i and VfV_f are at the initial and final positions.
  • Not converting eV to joules. When you mix units, convert: 1 eV = 1.60×10−191.60 \times 10^{-19} J.
  • Assuming every charge gains KE. Only charges moving “downhill” in potential (the way the field pushes them) speed up on their own.

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