The surface densities on the surfaces of two charged spherical conductors of radii $R_{1}$ and $R_{2}$ are equal. The ratio of electric intensities on their surfaces is:

R_{1}^{2} / R_{2}^{2}

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R_{2}^{2} / R_{1}^{2}

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R_{1} / R_{2}

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1 : 1

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Solution

The correct option is C $1 : 1$
Let the surface densities on the surfaces of two charged spherical conductors of radii $R_{1}$ and $R_{2}$ be $σ_{1}$ and $σ_{2}$ respectively.
But $σ_{1} = σ_{2}$ ...(given)
Electric field intensity on surface of spherical conductor is given by,

$E = \frac{Q}{4 π ϵ_{o} r^{2}} = \frac{σ}{ϵ_{o}}$

$∴$ Ratio of electric field intensities on their surfacesis given by,

$\frac{E_{1}}{E_{2}} = \frac{\frac{σ_{1}}{ϵ_{o}}}{\frac{σ_{2}}{ϵ_{o}}} = \frac{σ_{1}}{σ_{2}} = \frac{1}{1}$ .... $(σ_{1} = σ_{2})$

$∴ E_{1} : E_{2} = 1 : 1$

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