Listrik Statis • Part 2: Contoh Soal Gaya Listrik, Medan Listrik, Potensial Listrik

Main ideas / lessons conveyed

• The video is an instructional Grade 12 physics lesson about static electricity, focusing on:
  • Electric force between point charges
  • Electric field (field strength) produced by charges
  • Electric flux (mentioned as part of the overall topic, though not worked in detail here)
  • Electric potential and electric potential energy (worked in the second example)
• The lesson includes two worked example problems:
  1. Finding the net electric force on charge B due to two other charges arranged with right-triangle geometry.
  2. Finding a point where the electric field is zero created by two charges, then finding the electric potential at that point.

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Example 1: Electric force on charge B (three charges at triangle vertices)

Given

• Three point charges located at vertices of right triangle ABC, with a right angle at B
• Side lengths:
  • (AB = 3\text{ cm})
  • (BC = 4\text{ cm})
• Charges:
  • At A: (q_A = +6\,\mu\text{C})
  • At B: (q_B = -12\,\mu\text{C})
  • At C: (q_C = -8\,\mu\text{C})

Core concepts used

• Like charges repel; opposite charges attract.

• Coulomb’s law for the magnitude of force between point charges: [ F = k\frac{|q_1 q_2|}{r^2} ]

• Forces are combined vectorially because they act in different perpendicular directions.

• Since the angle at B is (90^\circ), the two force components on charge B are treated as perpendicular vectors.

Step-by-step methodology (as presented)

1. Draw the right triangle ABC with the right angle at B.
2. Convert distances to meters:
   • (AB = 3\times 10^{-2}\,\text{m})
   • (BC = 4\times 10^{-2}\,\text{m})
3. Determine directions of forces on charge B:
   • Force on B due to A: (q_A(+)) and (q_B(-)) ⇒ attraction
   • Force on B due to C: (q_C(-)) and (q_B(-)) ⇒ repulsion

4. Compute magnitudes using Coulomb’s law (magnitudes use absolute values):

   • [ F_{BA} = k\frac{|q_B q_A|}{(AB)^2} ]

   • [ F_{BC} = k\frac{|q_B q_C|}{(BC)^2} ]

5. Combine perpendicular components: [ F_{\text{net}}=\sqrt{F_{BA}^2+F_{BC}^2} ] (The transcript references a cosine form in general, but the calculation effectively uses the (90^\circ) right-triangle result.)

Result (as concluded in the subtitles)

• Net electric force on charge B:

  • [ F_{\text{net}} \approx 9.0\times 10^2\ \text{N} ]

• The spoken number is noisy, but the calculation ends at roughly 900 N.

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Example 2: Two charges—find point where electric field is zero, then electric potential there

Given

• Two point charges separated by 8 cm:
  • (q_A = +144\,\mu\text{C})
  • (q_B = +16\,\mu\text{C})
  • (AB = 8\text{ cm} = 0.08\text{ m})
• Tasks:
  1. Find a point C where the electric field strength is zero.
  2. Find the electric potential at that point.

Core concepts used

• Electric field magnitude from a point charge: [ E = k\frac{|q|}{r^2} ]

• Electric field direction:

  • For positive charges, the field points away from the charge.
• Condition for zero net electric field:
  • Field vectors must cancel (opposite directions with equal magnitudes).
• Electric potential is a scalar and adds directly: [ V = k\frac{q}{r} ] [ V_C = V_{CA} + V_{CB} ]

Step-by-step methodology (as presented)

Part (a): Locate where (E = 0)

• Reject the point between A and B:
  • Between two positive charges, both electric fields point the same way, so they cannot cancel.
• Assume the zero-field point C is to the right of B (outside the segment).
• Let:
  • Distance from B to C be (x)
  • Then distance from A to C is (0.08 + x)

• Set field magnitudes equal for cancellation: [ k\frac{q_A}{(0.08+x)^2} = k\frac{q_B}{x^2} ]

• Solve for (x):

  • [ x = 0.04\,\text{m} ]

  • So C is 4 cm to the right of B.

Part (b): Compute electric potential at C

• Add potentials (scalar addition): [ V_C = k\frac{q_A}{r_{AC}} + k\frac{q_B}{r_{BC}} ]

• Substitute distances:

  • (r_{AC} = 0.08 + 0.04 = 0.12\,\text{m})
  • (r_{BC} = 0.04\,\text{m})

• Subtitle result:

  • [ V_C \approx -7.2\times 10^6 ]

• Note: The negative sign appears inconsistent with both charges being stated as positive; the transcript is described as noisy/inconsistent, but the final numeric value is about (7.2\times 10^6).

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Speakers / sources featured

• Christian Sutantio — host/presenter
• Science Window Channel — YouTube channel/source of the lesson