Electrostatic Potential and Capacitance
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New answer posted
a month agoContributor-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 agoNew answer posted
a month agoContributor-Level 9
Magnitude of electric field is constant & the surface is equipotential.
New answer posted
a month agoContributor-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
New answer posted
a month agoContributor-Level 10
With the help of conservation of volume, we can write
With the help of conservation of charge, we can write
Q = 27 q.(2)
Potential energy of single drop = U1 =
Potential energy of bigger drop =
New answer posted
a month agoNew answer posted
a month agoContributor-Level 10
Let the charge of on each drop be q and radius of each drop be r.
When all drops are joined, radius,
Potential of the new drop,
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