Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by - Physics MCQ AIEEE 2006 2026 | ExamTest.live

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Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire, then in equilibrium, the ratio of the magnitude of the electric fields at the surface of spheres A and B is:

Solution & Step-by-step Explanation

When connected by a wire, the potentials of the two spheres become equal: V_{A} = V_{B} . \frac{1}{4 π ϵ _{0}} \frac{Q _{A}}{r _{A}} = \frac{1}{4 π ϵ _{0}} \frac{Q _{B}}{r _{B}} ⟹ \frac{Q _{A}}{r _{A}} = \frac{Q _{B}}{r _{B}} Electric field at the surface E = \frac{1}{4 π ϵ _{0}} \frac{Q}{r ^{2}} .Ratio of fields: \frac{E _{A}}{E _{B}} = \frac{Q _{A} / r _{A}^{2}}{Q _{B} / r _{B}^{2}} = \frac{Q _{A}}{Q _{B}} \times (\frac{r _{B}}{r _{A}})^{2} Since \frac{Q _{A}}{Q _{B}} = \frac{r _{A}}{r _{B}} : \frac{E _{A}}{E _{B}} = \frac{r _{A}}{r _{B}} \times \frac{r _{B}^{2}}{r _{A}^{2}} = \frac{r _{B}}{r _{A}} Given r_{A} = 1 mm, r_{B} = 2 mm . \frac{E _{A}}{E _{B}} = \frac{2}{1}

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Two spherical conductors A and B of radii 1 mm and 2 mm are separated by a distance of 5 cm and are uniformly charged. If the spheres are connected by a conducting wire, then in equilibrium, the ratio of the magnitude of the electric fields at the surface of spheres A and B is:

1 : 4

4 : 1

1 : 2

2 : 1

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