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General Physics Lesson 3-4
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Lesson 3: Electric Flux and Gauss's Law Example Problem 1:
A disk of radius 0.10 m is oriented with its area
Electric Flux (Φ) vector at 30° to a uniform electric field of
— it is the measure of flow of electric field 3
magnitude 𝐸 = 2. 0 × 10 𝑁/𝐶.
through a surface a. What is the electric flux through the disk?
2
— unit: 𝑁𝑚 /𝐶 (Newton-meter squared per b. What is the electric flux if the Area vector
coulomb) is parallel to the Electric Field?
— it is the product of the magnitude of the c. What is the electric flux if the Area vector
electric field and the surface area, A, is perpendicular to the Electric Field?
perpendicular to the field given by the
equation below: Answers:
Φ = 𝐸𝐴
॰ Φ - Electric flux
॰ 𝐸 - Electric Field
॰ 𝐴 - Area
— it is proportional to the number of electric
field lines penetrating some surface
— the field lines may make some angle θ that is
perpendicular to the surface
Φ = 𝐸𝐴 𝑐𝑜𝑠 θ
॰ Φ - Electric flux
॰ 𝐸 - Electric Field
॰ 𝐴 - Area
॰ θ - the angle between Electric Field and
the Area vector

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