12th Physics, Chapter 12th, Punjab text Book

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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