Physics Electrostatic Potential and Capacitance

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R
Raj Pandey

Contributor-Level 9

C = A B ? A ? B ?

Using De-Morgan Theorem

C = A B ? A ? B ?

C = B ( A + A ? ) ? = B ?

Therefore

 

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A
alok kumar singh

Contributor-Level 10

Time period of satellite

T = 2 π R 3 G M

= 2 π R 3 G d 4 3 π R 3

T = 3 π G d 3 π G d = T 2

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R
Raj Pandey

Contributor-Level 9

Electric field is always perpendicular to equipotential surface.

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A
alok kumar singh

Contributor-Level 10

No current through '  G  '

So potential difference across R is 2 V

i = 8 400 R = 2 8 * 400 = 100 Ω

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V
Vishal Baghel

Contributor-Level 10

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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A
alok kumar singh

Contributor-Level 10

The flux φ is calculated as:
φ = (2/5) * 4 * 10³ * 0.4 = 640 Nm²C? ¹

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A
alok kumar singh

Contributor-Level 10

Let the charge on C? be q µC. For the capacitor network:
q/C? = (C? * 10 - q) / C? ⇒ q/8 = (2 * 10 - q) / 2 ⇒ q = 16

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V
Vishal Baghel

Contributor-Level 10

Kindly consider the following figure

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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

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

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