A Geiger tube consists of two elements, a long metal cylindrical shell and a lon

Question

A Geiger tube consists of two elements, a long metal cylindrical shell and a long straight metal wire running down its central axis. Model the tube as if both the wire and cylinder are infinitely long. The central wire is positively charged and the outer cylinder is negatively charged. The potential difference between the wire and the cylinder is 0.99 kV. Suppose the cylinder in the Geiger tube has an inside diameter of 3.80 cm and the wire has a diameter of 0.464 mm. The cylinder is grounded so its potential is equal to zero.
(a) What is the radius of the equipotential surface that has a potential equal to 495 V? Is this surface closer to the wire or to the cylinder?

Explanation / Answer

   radius of the central wire   rw =   0.464/ 2   =   0.232   mm    distance between wire and the cylinder   d   =   ro - rw                                                                                  =   19   -   0.232                                                                            =   18.768  mm                                                                                                                                                  =   1.8768  cm    electric field strength      E   =   V / d                                                 =   0.99 kV / 1.8768 cm                                                 =   527.4    V/cm    a.   Distance of V1 =  495 V surface from cylinder          d1   =   V1 / E                =   495 / 527.4                =   0.938 cm          hence the radius   r1   =  ro - d1   =   1.9cm - 0.938cm                                                                 = 0.962cm

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