# 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?

Electric Field Intensity -

$\dpi{100} \vec{E}=\frac{\vec{F}}{q_{0}}=\frac{kQ}{r^{2}}$

- wherein

Capacitance of Conductor -

$\dpi{100} Q\propto V$

$\dpi{100} Q=CV$

- 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  $E_{B}= \frac{V}{R_{B}}$

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

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