Energía potencial eléctrica | Física | Khan Academy en Español

Main ideas and lessons

• Key confusion resolved: When two like charges are released, they separate and both gain kinetic energy (they speed up). The question is: where does that kinetic energy come from if energy is conserved?

  • The explanation is that the charges’ electrical potential energy decreases while their kinetic energy increases.
  • The total energy of the two-charge system remains constant; energy is transformed, not created.

• Electrical potential energy as a system property

  • Electrical potential energy is not associated with just one charge.
  • You need at least two charges to define electrical potential energy.
  • Therefore, the quantity computed by the formula is the potential energy of the system (not of (q_1) alone or (q_2) alone).

• Scalar nature

  • Electric potential energy is a scalar, so it:
    • has units of joules (SI)
    • has no direction
    • does not require vector decomposition into x/y components.

• Core formula for electric potential energy

  • The video uses the standard form (result stated without calculus): [ U = \frac{K\,q_1\,q_2}{r} ]

  • Distance rule: the separation (r) is not squared.

    • The video contrasts this with Coulomb’s law (where (r^2) appears) and says the integral in the derivation effectively changes (r^2) into (r).
  • Symbol note: potential energy is commonly denoted by (U); sometimes (E) appears with a subscript for clarification (e.g., electrical vs gravitational).

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Step-by-step methodology shown (worked example)

Example 1: Like charges repel (start at 3 cm, end at 12 cm)

Given

• Charges (q_1, q_2) (numerical values given):
  • (q_1 = 4\,\mu C = 4\times10^{-6}\,C)
  • (q_2 = 2\,\mu C = 2\times10^{-6}\,C)
• Mass of each charge:
  • (m_1 = m_2 = 1\,kg)
• Initial separation (center-to-center):
  • (r_i = 3\,cm = 0.03\,m)
• Final separation (center-to-center):
  • (r_f = 12\,cm = 0.12\,m)
• Constant:
  • (K = 9\times10^9)

Goal

• Find the speed of both charges when they reach (r_f).

Method using conservation of energy

1. Write energy conservation for the two-charge system
   • Total initial energy = total final energy.
   • System starts at rest ⇒ initial kinetic energy is 0.
2. Identify energies
   • Initial: only electrical potential energy (U_i)
   • Final: electrical potential energy (U_f) + kinetic energy of both charges

3. Compute initial potential energy [ U_i=\frac{Kq_1q_2}{r_i} ]

   • The video arrives at (U_i = 2.4\,J) after correct unit conversion. 4. Compute final potential energy [ U_f=\frac{Kq_1q_2}{r_f} ]

   • The video arrives at (U_f = 0.6\,J). 5. Set up conservation equation [ U_i = U_f + K_{\text{total}} ] [ 2.4 = 0.6 + K_{\text{total}} \Rightarrow K_{\text{total}}=1.8\,J ]

4. Kinetic energy must include both charges

   • Since both particles move: [ K_{\text{total}}=\frac12 m v_1^2+\frac12 m v_2^2 ]

   • With equal masses and equal speeds (argument based on Newton’s 3rd law / equal work and same mass):

     • (v_1=v_2=v)
     • [ K_{\text{total}}=\frac12(1)v^2+\frac12(1)v^2 = v^2 ]

5. Solve for speed [ v^2 = 1.8 \Rightarrow v=\sqrt{1.8}\approx 1.3\,m/s ]

Result

• Both charges move at about (1.3\,m/s) when separated by 12 cm.

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Example 2: Opposite charges attract (start at 12 cm, end at 3 cm; (q_2) becomes negative)

Change from Example 1

• The video considers what happens if one charge sign flips:
  • (q_2) becomes negative: (q_2 = -2\,\mu C)
• Release from:
  • (r_i = 12\,cm = 0.12\,m)
• Final separation:
  • (r_f = 3\,cm = 0.03\,m)
• Still starts from rest.

Updated method

1. Recompute potential energies with signs
   • Because (U=\dfrac{Kq_1q_2}{r}), flipping the sign of (q_2) makes (U) negative.
2. Initial potential energy becomes more “negative”
   • The video gets:
     • (U_i = -0.6\,J)
3. Final potential energy becomes even more negative
   • The video gets:
     • (U_f = -2.4\,J)

4. Use conservation of energy [ U_i = U_f + K_{\text{total}} ] [ -0.6 = -2.4 + K_{\text{total}} \Rightarrow K_{\text{total}}=1.8\,J ]

5. Solve for speed

   • Same kinetic-energy equation for equal masses:
     • (K_{\text{total}}=v^2)
   • [ v=\sqrt{1.8}\approx 1.3\,m/s ]

6. Important conceptual point

   • Having negative potential energy is allowed.
   • The system’s kinetic energy increases because the potential energy becomes even more negative (i.e., it “drops” further), supplying energy to motion.

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Final takeaway (stated summary)

• Electric potential energy formula: [ U = \frac{Kq_1q_2}{r} ]

• (r) is center-to-center separation and is not squared.

• (U) is a scalar, so it:
  • has no direction
  • its sign comes from the signs of (q_1) and (q_2)
• Conservation of energy applies even when (U) is negative:
  • the system can still convert electrical potential energy into kinetic energy.

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Speakers / sources featured

• Khan Academy en Español (educational video presenter) — no individual name is given in the subtitles.