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 condition, the ratio of the magnitude of the electric fields at the surface of spheres A and B is 

  • Option 1)

    1:4

  • Option 2)

    4:1

  • Option 3)

    1:2

  • Option 4)

    2:1

 

Answers (2)

As we learnt in

Electric Field Intensity -

\vec{E}=\frac{\vec{F}}{q_{0}}=\frac{kQ}{r^{2}}

- wherein

 

 

Capacitance of Conductor -

Q\propto V

Q=CV

- wherein

C - Capacity or capacitance of conductor 

V - Potential.

 

 When the spherical conductors are connected by a conducting wire, charge is redistributed and the spheres attain a common potentia  V.

\therefore\: \: \: Intensity \: E_{A}= \frac{1}{4\pi \varepsilon _{0}}\frac{Q_{A}}{R_{A}^{2}}

or\: \: \: \: E_{A}= \frac{1\times C_{A}V}{4\pi \varepsilon _{0}R_{A}^{2}}= \frac{\left ( 4\pi \varepsilon _{0}R_{A} \right )V}{4\pi \varepsilon _{0}R_{A}^{2}}= \frac{V}{R_{A}}

Similarly  E_{B}= \frac{V}{R_{B}}

\therefore \frac{E_{A}}{E_{B}}= \frac{R_{B}}{R_{A}}= \frac{2}{1}


Option 1)

1:4

Incorrect

Option 2)

4:1

Incorrect

Option 3)

1:2

Incorrect

Option 4)

2:1

Correct

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