The document discusses the application of Gauss's law to calculate electric field intensity in different scenarios. It first examines electric field intensity inside a hollow charged sphere, then calculates the electric field intensity due to an infinite sheet of charge and between two oppositely charged parallel plates. For each scenario, it outlines the 4 steps of using Gauss's law: 1) choosing a Gaussian surface, 2) determining the enclosed charge, 3) calculating the total flux, and 4) using Gauss's law to find the electric field intensity.

1. APPLICATION OF GAUSS’S LAW
Dr. Nadeem Khalid
AL-AQSA ACADEMY, LAHORE

2.  Intensity of field inside a Hollow Charged Sphere
• E = 0
 Electric intensity due to an infinite sheet of charge
• E =
𝜎
2ϵ0
 Electric intensity between to oppositely charged Parallel plates
• E =
𝜎
ϵ0
PLAN TO TALK
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3. • 4 steps
1. Gaussian Surface
2. Charged enclosed
3. Total flux
4. Electric intensity
ELECTRIC INTENSITY DUE TO AN
INFINITE SHEET OF CHARGE
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4. •Step 1
• Gaussian surface
ELECTRIC INTENSITY DUE TO AN
INFINITE SHEET OF CHARGE
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5. •Step 2
• Charged enclosed
• To find this first we know about concept of surface charge density(𝜎)
• 𝜎 =
𝑞
𝐴
……..(A)
• From equation …(A)
• q = 𝜎 𝐴
ELECTRIC INTENSITY DUE TO AN
INFINITE SHEET OF CHARGE
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6. •Step 3
• Total flux
1. Right end flat surface
2. Left end flat surface
3. Curved surface
 Right end flat surface
ELECTRIC INTENSITY DUE TO AN
INFINITE SHEET OF CHARGE
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7.  Left end flat surface
 Flux through curved surface
• 𝜑e3 = 0
 Total flux
 𝜑 = 2EA……….(1)
ELECTRIC INTENSITY DUE TO AN
INFINITE SHEET OF CHARGE
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8. •Step 4
• Calculation of electric intensity
• According to gauss’s law
• 𝜑e =
1
ϵ0
(q)…………(2)
• Using 1 & 2
• 2EA = 1
ϵ0
(q)
• 2EA = 1
ϵ0
(𝜎 𝐴)
•E = 𝜎
2ϵ0
ELECTRIC INTENSITY DUE TO AN
INFINITE SHEET OF CHARGE
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9. • 4 steps
1. Gaussian Surface
2. Charged enclosed
3. Total flux
4. Electric intensity
ELECTRIC INTENSITY BETWEEN
TWO OPPOSITELY CHARGED
PARALLEL PLATES
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10. •Step 1
• Gaussian surface
ELECTRIC INTENSITY BETWEEN
TWO OPPOSITELY CHARGED
PARALLEL PLATES
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11. •Step 2
• Charged enclosed
• To find this first we know about concept of surface charge density(𝜎)
• 𝜎 =
𝑞
𝐴
……..(A)
• From equation …(A)
• q = 𝜎 𝐴
ELECTRIC INTENSITY BETWEEN
TWO OPPOSITELY CHARGED
PARALLEL PLATES
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12. • Step 3
•Total flux
1. Flux through sides of box
2. Flux through upper surface of box
3. Flux through lower surface of box
• Flux through sides of box
• 𝜑 = 0
• Flux through upper surface of box
• 𝜑 = 0
ELECTRIC INTENSITY BETWEEN
TWO OPPOSITELY CHARGED
PARALLEL PLATES
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13. • Flux through lower surface of box
• Total flux
• 𝜑 = EA………(1)
ELECTRIC INTENSITY BETWEEN
TWO OPPOSITELY CHARGED
PARALLEL PLATES
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14. • Step 4
• Calculation of electric intensity
• 𝜑e =
1
ϵ0
(q)…………(2)
• Using 1 & 2
• EA = 1
ϵ0
(q)
• EA = 1
ϵ0
(𝜎 𝐴)
•E = 𝜎
ϵ0
ELECTRIC INTENSITY BETWEEN
TWO OPPOSITELY CHARGED
PARALLEL PLATES
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16. ANSWER : D
PRACTICE TEST
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17. ANSWER : C
PRACTICE TEST
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