Question:
Space between two concentric conducting spheres of radii a and b (b > a) is filled with a medium of resistivity \(\rho\). The resistance between the two spheres will be :
Space between two concentric conducting spheres of radii a and b (b > a) is filled with a medium of resistivity \(\rho\). The resistance between the two spheres will be :
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Treat the medium as thin concentric shells of thickness dr in series and integrate \(dR=\rho\,dr/(4\pi r^2)\).
Updated On: Nov 5, 2026
- \(\frac{\rho}{2\pi}\left(\frac1a+\frac1b\right)\)
- \(\frac{\rho}{4\pi}\left(\frac1a+\frac1b\right)\)
- \(\frac{\rho}{2\pi}\left(\frac1a-\frac1b\right)\)
- \(\frac{\rho}{4\pi}\left(\frac1a-\frac1b\right)\)
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The Correct Option is D
Solution and Explanation
Consider a thin spherical shell of radius r and thickness dr. Its resistance is \(dR=\rho\frac{dr}{4\pi r^2}\) (length dr, area \(4\pi r^2\)).
The shells are in series: \(R=\int_a^b\frac{\rho\,dr}{4\pi r^2}=\frac{\rho}{4\pi}\left[-\frac1r\right]_a^b=\frac{\rho}{4\pi}\left(\frac1a-\frac1b\right)\).
Answer: (4) \(\frac{\rho}{4\pi}\left(\frac1a-\frac1b\right)\)
The shells are in series: \(R=\int_a^b\frac{\rho\,dr}{4\pi r^2}=\frac{\rho}{4\pi}\left[-\frac1r\right]_a^b=\frac{\rho}{4\pi}\left(\frac1a-\frac1b\right)\).
Answer: (4) \(\frac{\rho}{4\pi}\left(\frac1a-\frac1b\right)\)
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