Verified Solution • 2025–26 Session moderate

QJEEM22Phfc5e: JEE Main 2022 Physics - Class 12 JEEM22Phfc5e Solution: Capacitance of an isolated conducting sphere of radius R 1…

Capacitance of an isolated conducting sphere of radius R1 becomes n times when it is enclosed by a concentric conducting sphere of radius R2 connected to earth. The ratio of their radii (R2R1)\left( {{{{R_2}} \over {{R_1}}}} \right) is :

Multiple Choice Options

  • A nn−1{n \over {n - 1}}
  • B 2n2n+1{{2n} \over {2n + 1}}
  • C n+1n{{n + 1} \over n}
  • D 2n+1n{{2n + 1} \over n}

Step-by-Step Verified Solution

Initially =C0=4πε0R1= {C_0} = 4\pi {\varepsilon _0}{R_1}

Finally 4πε0R1R2R2−R1=nC0=4πε0nR1{{4\pi {\varepsilon _0}{R_1}{R_2}} \over {{R_2} - {R_1}}} = n{C_0} = 4\pi {\varepsilon _0}n{R_1}

⇒\Rightarrow R2R2−R1=n{{{R_2}} \over {{R_2} - {R_1}}} = n

⇒\Rightarrow 1−R1R2=1n1 - {{{R_1}} \over {{R_2}}} = {1 \over n}

⇒\Rightarrow R1R2=n−1n{{{R_1}} \over {{R_2}}} = {{n - 1} \over n}

⇒\Rightarrow R2R1=nn−1{{{R_2}} \over {{R_1}}} = {n \over {n - 1}}

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