The 3.20 - km - longSLAC produces a beam of 50 GeV electrons. If there are 15,000 accelerating tubes, what average voltage must be across the gaps between them to achieve this energy?

Voltage is the pressure from the power source of an electrical circuit that allows charged electrons (currents) to pass through conductive loops and allow them to do their jobs. B. Make the light shine.

Length of the accelerator, l = 3.20 km

Number of accelerating tubes,\(n = 15000\)

Energy of the beam,\(E = 50.0\;{\rm{GeV}}\)

The general voltage among the 15,000 tubes can be obtained using the following formula \({\rm{P}} \cdot {\rm{E = e}}{{\rm{V}}_{\rm{t}}}\), where P. E is the energy of the beam of the electrons and V_t is the total voltage.

By dividing the total voltage of the 15,000 tubes, you can get the voltage of the gap between them:

\(\begin{array}{c}{{\rm{V}}_{\rm{t}}}{\rm{ = }}\frac{{{\rm{P}}{\rm{.E}}}}{{\rm{e}}}\\{\rm{ = }}\frac{{{\rm{50}}\;{\rm{GeV}}}}{{\rm{e}}}\\{\rm{ = 50}}\;{\rm{GV}}\end{array}\)

\(\begin{array}{c}{\rm{V = }}\frac{{{\rm{50}}\;{\rm{GV}}}}{{{\rm{15,000}}}}\\{\rm{ = 3}}{\rm{.33 \times 1}}{{\rm{0}}^{\rm{6}}}{\rm{\;V}}\end{array}\)

Hence, the voltage across the gap between the tubes 3.33MV.