Physics Electrostatic Potential and Capacitance

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

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) = CV²/ (2k)

U_f = (14 * 10? ¹² * 12²) / (2 * 7) = 144 pJ

Mechanical energy available for oscillation

New answer posted

a month ago

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alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

0 Follower 1 View

A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

E = E? (1 − ax²)

F = qE?
acceleration = F/m = (qE? /m) (1 - ax²) = v (dv/dx)
(qE? /m) ∫? (1 - ax²)dx = ∫? vdv; (qE? /M) (x - ax³/3) = 0
x (1 - ax²/3) = 0; x = 0 & x = √ (3/a)

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alok kumar singh

Contributor-Level 10

By conservation of energy
(K.E. + P.E.)? = (K.E. + P.E.)?
0 + (1/ (4πε? a³) * 2P * P = 2 * (1/2)mv² + 0
v = √ (p² / (2πε? ma³) = (P/a) * √ (1 / (2πε? ma)

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

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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