Question

A counter consists of a cylindrical cathode of radius 1cm and an anode wire of radius 0.01cm which is placed along the axis of the cathode. A voltage of 2.3kV is applied between the cathode and anode. The electric field on the anode surface must be:
(A) $2.3\times {{10}^{5}}$ V/m
(B) $5\times {{10}^{6}}$V/m
(C) $4.6\times {{10}^{5}}$V/m
(D) $2.5\times {{10}^{6}}$V/m

Answer 1

We have a cylindrical shape and given outer radius and inner radius. Geiger counters are used to detect radiation viz beta and alpha particles. The counter consists of a tube filled with an inert gas that becomes conductive of electricity when it is impacted by a high-energy particle.

Complete step by step answer:
The radius of the outer cylinder is ${{R}_{2}}=1$cm
The radius of the inner wire cylinder ${{R}_{1}}=0.01$cm
For cylindrical capacitor capacitance is given by the formula, $C=\dfrac{2\pi {{\varepsilon }_{0}}L}{\ln (\dfrac{{{R}_{2}}}{{{R}_{1}}})}$
Now given in the question the value of potential difference across the ends is 2.3kV viz 2300 V
Using the formula to calculate the charge, Q=CV
Q= $CV=\dfrac{2\pi {{\varepsilon }_{0}}LV}{\ln (\dfrac{{{R}_{2}}}{{{R}_{1}}})}$
Now let us find the charge density that is charge per unit length denoted by $\lambda$
$\lambda$= $=\dfrac{Q}{L}$
$=\dfrac{2\pi {{\varepsilon }_{0}}V}{\ln (\dfrac{{{R}_{2}}}{{{R}_{1}}})}$
We know electric field due to a rod or wire in the shape of cylindrical is given by $E=\dfrac{\lambda }{2\pi {{\varepsilon }_{0}}{{R}_{1}}}$
Substituting the values,
$=\dfrac{V}{{{R}_{1}}\ln (\dfrac{{{R}_{2}}}{{{R}_{1}}})}$
$=\dfrac{2300}{(0.01\times {{10}^{-2}})\ln (100)}$
= $5\times {{10}^{6}}V/m$
This matches with the option (B), hence, the correct option is (B).

Note: While substituting the values we should keep in mind that all the units must be in standard SI. The voltage was given in kilo Volt, w converted into volt. The counter has a cylindrical capacitor and so formula for cylindrical capacitance is used.