Electrostatic Potential and Capacitance

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A
alok kumar singh

Contributor-Level 10

Kindly go through the solution

 

New answer posted

a month ago

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A
alok kumar singh

Contributor-Level 10

Uinitial = k(4q)(q)/(d/2) + k(q)(-q)/(d/2)
⇒ 6kq²/d
⇒ Ufinal = k(4q)(q)/((3d/2)) + k(q)(-q)/(d/2)
⇒ 2kq²/3d
⇒ ΔU = (2/3 - 6)kq²/d = -16kq²/3d

New question posted

a month ago

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New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Given:   C 1 + C 2 = 10 μ F

4 1 2 C 1 V 2 = 1 2 C 2 V 2 4 C 1 = C 2

From equation (i) and (ii)

C 1 = 2 μ F C 2 = 8 μ F If they are in series

C eq   = C 1 C 2 C 1 + C 2 = 1.6 μ F

New answer posted

a month ago

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R
Raj Pandey

Contributor-Level 9

Magnitude of electric field is constant & the surface is equipotential.

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A
alok kumar singh

Contributor-Level 10

According to Question, we can write
E = λ/ (2πε? r) ⇒ σ = ε? E = λ/ (2πr)
⇒ σ = 8*10? / (2*3.14*3) = 0.424 nCm? ²

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a month ago

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V
Vishal Baghel

Contributor-Level 10

With the help of conservation of volume, we can write

2 7 * 4 3 π r 3 = 4 3 π R 3 R = 3 r . . . . . . . ( 1 )

With the help of conservation of charge, we can write

Q = 27 q.(2)

Potential energy of single drop = U1 = q 2 8 π ε 0 r

Potential energy of bigger drop = U 2 = Q 2 8 π ε 0 R = 2 7 * 2 7 * q 2 8 π ε 0 ( 3 r ) = 2 4 3 ( q 2 8 π ε 0 r ) = 2 4 3 U 1

U 2 U 1 = 2 4 3

 

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

F n e t = 2 k e e ( d 2 + x 2 ) s i n θ = ( 2 k e 2 ( d 2 + 2 ) ) θ

[for small displacement i.e, x is small so|, sin   tan ]

= 2 k e 2 ( d 2 + x 2 ) . x d = 2 . 1 4 π ε 0 q 2 x d 3   [q = e]

= q 2 2 π ε 0 d 3 x = m ω 2 x

ω = q 2 2 π ε 0 d 3 m

New answer posted

a month ago

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V
Vishal Baghel

Contributor-Level 10

Let the charge of on each drop be q and radius of each drop be r.

k q r = 2

When all drops are joined, radius,

r ' = ( 5 1 2 ) 1 3 r = 8 r

Potential of the new drop,

V = k . 5 1 2 q 8 r = 6 4 k q r = 1 2 8 V

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