JEE Main & Advanced Physics Electrostatics & Capacitance JEE PYQ-Electrostatics and Capacitance

  • question_answer
    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, the in equilibrium condition, the ratio of the magnitude of the electric fields at the surfaces of spheres A and B is                                    [AIEEE 2006]

    A) 4 : 1

    B) 1 : 2

    C) 2 : 1

    D) 1 : 4

    Correct Answer: A

    Solution :

    [a] By connecting two surfaces with the help of a conductor wire, we make the potential of each of the surfaces equal remembering that electric field at a point is inverse proportional to the distance.
    When the two conducting spheres are connected by a conducting wire, charge will flow from one sphere (having higher potential) to other (having lower potential) till both acquire the same potential.
    \[{{V}_{1}}=\frac{kq}{{{r}_{1}}},{{V}_{2}}=\frac{kq}{{{r}_{2}}}\Rightarrow {{v}_{1}}={{v}_{2}}\]
    (according to question)
    \[\Rightarrow \]   \[{{r}_{2}}=2{{r}_{1}},E=\frac{1}{4\pi {{\varepsilon }_{0}}}.\frac{q}{{{r}^{2}}}\]
    So,       \[\frac{{{E}_{1}}}{{{E}_{2}}}={{\left( \frac{{{r}_{2}}}{{{r}_{1}}} \right)}^{2}}=4:1\]


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