Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
JEE MainPhysicsElectrostatics

Two concentric circular arcs are fixed in space. The inner arc is a semicircle of radius R with a uniform linear charge density + . The outer arc is a quarter-circle of radius 2R with a uniform linear charge density - . A particle of mass m and positive charge +q is released from rest at the common centre of the arcs. The kinetic energy of the particle when it reaches a point very far away (infinity) is:

Options

  1. A0
  2. B3q 8 ₀
  3. Cq 4 ₀
  4. Dq 8 ₀

Correct answer

D. q 8 ₀

Step-by-step solution

The electric potential at the centre due to a circular arc of radius r subtending an angle and having linear charge density is V = 1 4 ₀ (r ) r = 4 ₀ . For the inner semicircle ( = ), the potential at the centre is: V₁ = (+ )( ) 4 ₀ = 4 ₀ For the outer quarter-circle ( = /2 ), the potential at the centre is: V₂ = (- )( /2) 4 ₀ = - 8 ₀ The net electric potential at the centre is: V_ net = V₁ + V₂ = 4 ₀ - 8 ₀ = 8 ₀ By conservation of mechanical energy, the kinetic energy of the particle at infinity equals its initial

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 JEE Main PYQs