Space between two concentric conducting spher | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) Consider a spherical shell of radius $x$ and thickness $dx$ between the two spheres.The area of this shell is $A = 4\pi x^2$.The resistance $dR$ o

Vedclass · Accepted Answer

$\frac{\rho}{4\pi} \left( \frac{1}{a} - \frac{1}{b} \right)$

Vedclass · Answer

(D) Consider a spherical shell of radius $x$ and thickness $dx$ between the two spheres.The area of this shell is $A = 4\pi x^2$.The resistance $dR$ of this thin shell is given by $dR = \rho \frac{dx}{A} = \rho \frac{dx}{4\pi x^2}$.To find the total resistance $R$ between the spheres,we integrate from $r = a$ to $r = b$:$R = \int_a^b \frac{\rho}{4\pi x^2} dx$.$R = \frac{\rho}{4\pi} \int_a^b x^{-2} dx$.$R = \frac{\rho}{4\pi} \left[ -\frac{1}{x} \right]_a^b$.$R = \frac{\rho}{4\pi} \left( -\frac{1}{b} - (-\frac{1}{a}) \right) = \frac{\rho}{4\pi} \left( \frac{1}{a} - \frac{1}{b} \right)$.

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

Similar Questions

$n$ conducting wires of same dimensions but having resistivities $1, 2, 3, . . . , n$ are connected in series. The equivalent resistivity of the combination is

The dimensions of resistivity in terms of $M, L, T$ and $Q$,where $Q$ stands for the dimensions of charge,is:

If copper and silicon are cooled from $300 \, K$ to $60 \, K$,their resistivity will:

$A$ metal rod of length $10 \ cm$ and a rectangular cross-section of $1 \ cm \times 0.5 \ cm$ is connected to a battery across opposite faces. The resistance will be . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module