Two charged spherical conductors of radius R 1 and R 2 are connected by a wire. Then the ratio of surface charge densities of the spheres ( \sigma 1 / \sigma 2 ) is :

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Electrostatics question

2021 · Q151

Electrostatics question

2021 · Q151

View written solutionFree

Correct answer: C

NEET 2021 Physics - Electrostatics Question 18 English Explanation

{Q_1} = {{\sum Q } \over {{R_1} + {R_2}}} \times {R_1}

{Q_2} = {{\sum Q } \over {{R_1} + {R_2}}} \times {R_2}

{\sigma _1} = {{{Q_1}} \over {4\pi R_1^2}} = {{\sum Q } \over {{R_1} + {R_2}}} \times {{{R_1}} \over {4\pi R_1^2}} \propto {1 \over {{R_1}}}

{\sigma _2} = {{{Q_2}} \over {4\pi R_2^2}} = {{\sum Q } \over {{R_1} + {R_2}}} \times {{{R_2}} \over {4\pi R_2^2}} \propto {1 \over {{R_2}}}

{{{\sigma _1}} \over {{\sigma _2}}} = {{{R_2}} \over {{R_1}}}

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Sign in

Electrostatics question

2021 · Q151

Electrostatics question

2021 · Q151

View written solutionFree

Correct answer: C

{Q_1} = {{\sum Q } \over {{R_1} + {R_2}}} \times {R_1}

{Q_2} = {{\sum Q } \over {{R_1} + {R_2}}} \times {R_2}

{\sigma _1} = {{{Q_1}} \over {4\pi R_1^2}} = {{\sum Q } \over {{R_1} + {R_2}}} \times {{{R_1}} \over {4\pi R_1^2}} \propto {1 \over {{R_1}}}

{\sigma _2} = {{{Q_2}} \over {4\pi R_2^2}} = {{\sum Q } \over {{R_1} + {R_2}}} \times {{{R_2}} \over {4\pi R_2^2}} \propto {1 \over {{R_2}}}

{{{\sigma _1}} \over {{\sigma _2}}} = {{{R_2}} \over {{R_1}}}