Listrik Statis • Part 3: Contoh Soal Fluks Listrik, Energi Potensial Listrik

Main ideas / lessons conveyed

The video continues Grade 12 static electricity example problems, focusing on:

• Electric potential and work required to move a charge in an electrostatic field.
• Electric flux through a surface in a uniform electric field, including how flux depends on the angle between the field and the surface normal.

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Example 1: Work required to move a charge toward/inside a charged conductor sphere

Given

A charged conductor sphere has:

• Total charge: [ Q = 200\,\mu C = 2.0 \times 10^{-4}\,C ]

• Radius: [ r = 40\,cm = 0.4\,m ]

A test charge is moved:

• [ q = 1\,\mu C = 1.0 \times 10^{-6}\,C ]

Points:

• Point A is 10 cm from the center: [ r_A = 0.1\,m ] (this is inside the sphere since (0.1 < 0.4))

• Point B is 10 cm from the surface: [ r_B = r + 0.1 = 0.4 + 0.1 = 0.5\,m ] (this is outside the sphere)

Method / steps (as presented)

Use: [ W_{AB} = q\left(V_B - V_A\right) ]

Find electric potential at each point.

Inside a conductor sphere (point A)

• Inside a conductor, the potential is constant.
• The transcript emphasizes using the correct radius-distance relationship (to avoid confusion).

Potential: [ V_A = k\frac{Q}{r} ]

Substituting:

• (k = 9 \times 10^9)
• (Q = 200 \times 10^{-6})
• (r = 0.4)

Result stated: [ V_A \approx 4.5 \times 10^6 \,V ]

Outside a conductor sphere (point B)

Potential depends on distance from the center: [ V_B = k\frac{Q}{r_B} ]

With (r_B = 0.5\,m), the result stated is: [ V_B \approx 3.6 \times 10^6 \,V ]

Compute work

[ W_{AB} = q(V_B - V_A) ]

With (q = 10^{-6}\,C), the transcript’s final result: [ W_{AB} \approx 0.9\,J ]

Key takeaway

• Electric potential inside a conductor is constant (set by the sphere’s radius).
• Outside, potential depends on distance from the center.
• Work to move a test charge is always: [ W = q\Delta V ]

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Example 2: Electric flux through a tilted sheet in a uniform electric field

Given

A rectangular paper surface:

• Length (= 30\,cm = 0.3\,m)
• Width (= 20\,cm = 0.2\,m)

Area: [ A = 0.3 \times 0.2 = 0.06\,m^2 ]

Uniform electric field magnitude:

• (E = 12\,N/C)

The paper is rotated by 60° relative to the direction of the electric field.

Method / steps (as presented)

Electric flux: [ \Phi = EA\cos\alpha ] where:

• (\alpha) is the angle between the electric field and the surface normal.

The transcript explains angle conversion:

• If the sheet makes (60^\circ) with the electric field direction, then the angle with the normal is: [ \alpha = 30^\circ ]

Warning emphasized: use the angle from the field to the normal in the cosine term (not necessarily the (60^\circ) relative to the surface).

Compute:

• (E = 12)
• (A = 0.3 \times 0.2)
• (\cos 30^\circ = \sqrt{3}/2)

The transcript arrives at an answer in terms of (\sqrt{3}), with flux expressed in weber (Wb) (though the transcription is messy, the calculation follows (EA\cos\alpha)).

Key takeaway

• Electric flux depends on orientation: [ \Phi \propto \cos\alpha ] where (\alpha) is measured between the electric field and the normal.

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Sources / speakers

• Christian Sutantio (host/presenter)
• “General Science Window” channel (source/channel mentioned by the speaker)