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Class Location
Online class via Zoom
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Tutor's Home
Years of Experience in Class 12 Tuition
5
Board
CBSE
Subjects taught
Physics, English, Mathematics, Chemistry
Taught in School or College
No
Answered on 18/10/2019 Learn CBSE - Class 12/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
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Here, C₁ = 600pF , V₁ = 20pV , C₂ = 600pF and V₂ = 0
On connecting charged capacitor to uncharged capacitor , the common potential V across the capacitors is given by
V = (C₁V₁ + C₂V₂)/(C₁ + C₂)
= (600 × 10⁻¹² × 200 + 600 × 10⁻¹² × 0)/(600 + 600) × 10⁻¹²
= 100V
now, energy stored in capacitors before connection is
Ui = 1/2 C₁V₁² + 1/2 C₂V₂²
= 1/2 × 600 × 10⁻¹² × (200)² + 1/2 × 600 × 10⁻¹² × 0
= 12 × 10⁻⁶ J = 12μJ
And energy stored in capacitors after connection is
Uf = 1/2 (C₁ + C₂)V²
= 1/2 × (600 + 600) × 10⁻¹² × (100)²
= 1/2 × 1200 × 10⁻⁸ J
= 6 × 10⁻⁶ J
= 6μJ
Hence , energy lost in the process is ∆U = Uf - Ui
= 6μJ - 12μJ = -6μJ [ negative sign indicates energy is lost.
Like 0 
Answers 8 
Comments Answered on 12/10/2019 Learn CBSE - Class 12/Chemistry/Unit IX: Coordination Compounds/NCERT Solutions/Intext Exercise 9.1
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Like 0 Answers 6 Comments Ask a Question
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Class Location
Online class via Zoom
Student's Home
Tutor's Home
Years of Experience in Class 12 Tuition
5
Board
CBSE
Subjects taught
Physics, English, Mathematics, Chemistry
Taught in School or College
No
Answered on 18/10/2019 Learn CBSE - Class 12/Physics/Unit 1-Electrostatics/ELECTROSTATIC POTENTIAL AND CAPACITANCE/NCERT Solutions/Exercise 2
Ask a Question
Post a Lesson
Here, C₁ = 600pF , V₁ = 20pV , C₂ = 600pF and V₂ = 0
On connecting charged capacitor to uncharged capacitor , the common potential V across the capacitors is given by
V = (C₁V₁ + C₂V₂)/(C₁ + C₂)
= (600 × 10⁻¹² × 200 + 600 × 10⁻¹² × 0)/(600 + 600) × 10⁻¹²
= 100V
now, energy stored in capacitors before connection is
Ui = 1/2 C₁V₁² + 1/2 C₂V₂²
= 1/2 × 600 × 10⁻¹² × (200)² + 1/2 × 600 × 10⁻¹² × 0
= 12 × 10⁻⁶ J = 12μJ
And energy stored in capacitors after connection is
Uf = 1/2 (C₁ + C₂)V²
= 1/2 × (600 + 600) × 10⁻¹² × (100)²
= 1/2 × 1200 × 10⁻⁸ J
= 6 × 10⁻⁶ J
= 6μJ
Hence , energy lost in the process is ∆U = Uf - Ui
= 6μJ - 12μJ = -6μJ [ negative sign indicates energy is lost.
Like 0 Answers 8 Comments Answered on 12/10/2019 Learn CBSE - Class 12/Chemistry/Unit IX: Coordination Compounds/NCERT Solutions/Intext Exercise 9.1
Ask a Question
Post a Lesson
Like 0 Answers 6 Comments Ask a Question
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