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

Hint: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.