Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
99 Percentile Qs Bank for JEE MainPhysicsElectrostatics

A conducting soap bubble having a radius a , thickness t < < a is charged to a potential V . If now, the bubble collapses and forms a droplet, then the potential of the droplet is

Options

  1. AV droplet = V a 3 t 1 3
  2. BV droplet = V a t 1 3
  3. CV droplet = V 3 a t 1 3
  4. DV droplet = V 3 a 4 t 1 3

Correct answer

A. V droplet = V a 3 t 1 3

Step-by-step solution

Let q be the charge on the bubble, then V = Kq a (Here K = 1 4 π ε 0 ) ∴ q = Va K Let after collapsing, the radius of droplet becomes R, then equating the volume, we have 4 π a 2 t = 4 3 π R 3 ∴ R = 3 a 2 t 1 / 3 Now, potential of droplet will be V ′ = Kq R Substituting the values, we have V ′ = K Va K 3 a 2 t 1 / 3 or V ′ = V a 3 t 1 / 3

Practice Electrostatics on Quantrex Academy →

More from Electrostatics

Two charges Q₁ = q and Q₂ = mq are placed at the points P₁(a, b) and P₂(ma, mb) , respectively, in the XY plane, where a, b 0 and m 0, 1 . If V₁ is the potential at a point in the 2026Consider an electric dipole comprising two charges +q and -q each with mass m , separated by a fixed distance d and initially at rest with its dipole moment pointing along i . A un 2026Two point charges q₁=3 , C and q₂=-4 , C are placed at points (2 i +3 j +3 k ) and ( i + j + k ) respectively. Force on charge q₂ is ________ N. ( Take 1 4 ₀ = 9 10^9 SI Units ) 2026The electric potential as a function of x, y is given by V = 5(x^2 - y^2) V. The electric field at a point (2, 3) m is __________ V/m. 2026A thin half ring of radius 35 cm is uniformly charged with a total charge of Q coulomb. If the magnitude of the electric field at centre of the half ring is 100 V/m, then the value 2026A three coulomb charge moves from the point (0, -2, -5) to the point (5, 1, 2) in an electric field expressed as E = 2x i + 3y^2 j + 4 k N/C. The work done in moving the charge is 2026A particle of charge q and mass m is projected from origin with an initial velocity v = ( v₀ 2 x + v₀ 2 y ) . There exists a uniform magnetic field B = B₀ z and a space varying ele 2026A rigid dipole undergoes a simple harmonic motion about its centre in the presence of an electric field E ₁=E₀ x . If another electric field E ₂=2E₀( y + z ) is introduced to the s 2026 Full Electrostatics list All 99 Percentile Qs Bank for JEE Main PYQs