Electrostatics

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New answer posted•a year ago
S
a year ago
Kindly go thorugh the solution

 

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New answer posted•a year ago
S
a year ago
Kindly go through the solution

 

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New answer posted•a year ago
S
a year ago

For a conducting sphere, the electric field E = σ/ε? and potential V = σR/ε?
When two spheres are connected by a wire, their potentials become equal: V? = V?
σ? R? /ε? = σ? R? /ε?
σ? R? = σ? R? ⇒ σ? /σ? = R? /R?

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New answer posted•a year ago
S
a year ago

F = (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + (qQ)/ (4πε? r? ²) + .

⇒ F = (qQ)/ (4πε? ) [1/1² + 1/2² + 1/4² + 1/8² + .]

⇒ F = (qQ)/ (4πε? ) [1/ (1 - 1/4)]

⇒ F = (qQ)/ (4πε? ) [4/3] = 10? × 1 × 9 × 10? × (4/3) = 12 × 10³ N

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New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Initial charge Q = CV = 14 × 10? ¹² × 12 = 168 × 10? ¹² C

Initial energy U_in = ½ CV² = ½ (14 × 10? ¹²) × 12² = 1008 pJ

When the battery is disconnected and a dielectric (k=7) is inserted, the new capacitance is C' = kC.

The charge Q remains constant.

Final energy U_f = Q²/2C' = Q²/ (2kC) = (CV)²/ (2kC)

...Read more

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New answer posted•a year ago
New answer posted•a year ago
New answer posted•a year ago
S
a year ago

Q? = CV?
U? = ½CV? ²
q? = (q? C)/ (C + C/2) = 2q? /3

q? = (q? (C/2)/ (C + C/2) = q? /3; Uf = q? ²/2C + q? ²/2 (C/2)
= (4q? ²/9x2C) + (q? ²/9C)
= (6q? ²/18C) = q? ²/3C = CV? ²/3

Energy loss in the process = U? – Uf
= ½CV? ² - CV? ²/3
= CV? ²/6

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