Relativistic Energy-mass relation If i charge an object and place it in an electric field does that mean its mass increases according to the equation $E=mc^2$? Or is the total energy in relativistic mass energy relation different? What kind of energy is this?
 A: The full relativistic equation of a particle's energy is:
\begin{equation}
E^2=p^2c^2+m_0^2c^4
\end{equation}
where $m_0$ is the rest mass of the particle, $p$ is its linear momentum.
When a particle is charged and placed in an electric field its rest mass $(m_0)$ does not change. Due to the motion induced by the external electric field momentum of the particle changes thereby changing the total energy.
$E_0=m_0c^2$ is the rest energy of the particle. This is not kinetic energy. This energy is a consequence of assembling the particle from nothing.
In the equation $E^2=p^2c^2+m_0^2c^4$, if we put $p=0$ (obviously no external force is present then) then we arrive at $E=E_0=m_0c^2$.
So obviously placing a charged particle in an electric field doesn't effect its mass.
But which part of the total energy changes? $E_0=m_0c^2$ can not. The kinetic energy changes.
If $K$ is the kinetic energy, $E=m_0c^2+K$.
And one can show, $K=\gamma m_0c^2-m_0c^2=m_0c^2(\gamma-1)$, where $\gamma=\frac{1}{\sqrt{1-\frac{v^2}{c^2}}}$ ($v$ is particle's velocity).
Also, it can be shown, when particle is moving very slowly $(v\lt\lt c)$, $K\approx\frac{1}{2}m_0v^2$.
A: You're confusing the term Mass with the term relativistic mass. Mass-or rest mass, is an inherent property of a body this is termed as $m_o$. Meanwhile what you're talking about is relativistic mass, which is a fictitious concept. The relativistic mass $m$ is given as-
$$m=\gamma m_o=\frac{m_o}{\sqrt{1-\frac{v^2}{c^2}}}$$
The actual equation is not $E=mc^2$, rather it is $E=\sqrt{(m_oc^2)^2+(m_ovc)^2}$. This would mean the energy term will change, because a charged body would move inside an electric field, however the mass, which is $m_o$ will not vary.
