Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202616 April 2026Evening ShiftPhysicsElectrostaticsActual

Two spherical hollow spheres of radii ' R₁ ' and ' R₂ ' are charged with same charge. Let ' ₁ ' and ' ₂ ' are their respective charge densities, then the ratio ₁ to ₂ is

Options

  1. AR₁ : R₂
  2. BR₂ : R₁
  3. CR₂^2 : R₁^2
  4. DR₁^2 : R₂^2

Correct answer

C. R₂^2 : R₁^2

Step-by-step solution

Let the charge on both the hollow spheres be Q . The surface charge density of a sphere of radius R is given by = Q 4 R^2 . For the first sphere of radius R₁ , the charge density is ₁ = Q 4 R₁^2 . For the second sphere of radius R₂ , the charge density is ₂ = Q 4 R₂^2 . Taking the ratio of their charge densities: ₁ ₂ = Q 4 R₁^2 Q 4 R₂^2 = R₂^2 R₁^2 Thus, the ratio ₁ : ₂ is R₂^2 : R₁^2 . Answer: R₂^2 : R₁^2

Practice Electrostatics on Quantrex Academy →

More from Electrostatics

A charge of 5 C is placed at the centre of a spherical shell S₁ of radius 10 cm. Now this system is enclosed inside another spherical shell S₂ of radius 20 cm. The ratio of the ele 2026Two point charges P = +25 C and Q = -16 C are placed 5 cm apart. Find the position of the point at which the resultant electric field is zero: 2026Bodies P, Q, R, S are labelled as having charges Q_P = 0.5 10⁻¹⁹C , Q_Q = 0.7 10⁻¹⁹C , Q_R = 2.1 10⁻¹⁹C , Q_S = 4.8 10⁻¹⁹C respectively. Select the body having the correct charge. 2026Uniform electric field of 5 10^3 ,NC⁻¹ is maintained in the positive Y direction. Now a point charge of 2 10⁻⁴C at rest is released from the origin. The kinetic energy attained by 2026A pith ball of mass 'm' gram and charge 'Q' is suspended using a mass less silk thread near a large charged conducting metal sheet of area 'A' and surface charged density ' '. If t 2026In a uniform electric field 10^6 N C⁻¹ an electric dipole of length 4 cm is placed with its axis making an angle 60^ with the electric field. If the dipole experiences a torque of 2026An electric field and magnetic field 1.8 10^4 Vm⁻¹ and 6 10⁻³ T respectively are applied simultaneously on an electron beam such that path of the beam remains undeviated, then the 2026A uniform electric field E = 3 i + 6 j + k passes through a closed cuboidal surface. One face of the cuboid has an area 4 m^2 and an outward unit normal given by 2 i +2 j +3 k 17 . 2026 Full Electrostatics list All MHT CET PYQs