Electrostatic Energy
Fondamental : Potential energy of a punctual charge q located at the point M under a potential V(M)
A particle of charge \(q\) is placed in an area of space where an electrostatic field \(\vec E\) reigns. It derives from the potential as such :
\(\vec E = - \overrightarrow {grad\;} V\)
The charge is subject to the force :
\(\vec f =q\vec E=- \overrightarrow {grad\;} (qV)\)
Hence the potential energy of the charge is :
\(E_p=qV\)
Attention : Potential energy of a punctual charge q located at the point M under a potential V(M)
\(E_p=qV\)
Exemple : Linear electron accelerator
Electrons are accelerated in a straight line (in empty space) from a metal electrode which has a potential equal to zero to a metal electrode which has a potential \(U>0\).
The first electrode is slightly heated, which allows the electrons to come out of it by thermoelectric effect.
Their initial velocity is approximately equal to zero (\(v_0\approx 0\)).
How can the velocity of the electrons \(v_1\) when they arrive to the second electrode be determined ?
The mechanical energy conservation of an electron can be used.
When the electron has the potential of the second electrode, this energy can be expressed by :
\(E_m=\frac {1}{2}mv_1^2-eU\)
Initially, the velocity and the potential are equal to zero.
Finally, the mechanical energy is equal to zero.
Hence, the velocity \(v_1\) of an electron when it reaches the second electrode verifies :
\(\frac {1}{2}mv_1^2-eU=0\)
So :
\(v_1=\sqrt{\frac {2eU}{m}}\)