Rankers Physics
Topic: Electrostatics
Subtopic: Properties of Charges

In relativistic mechanics \(m=\frac{m_{0}}{\sqrt{\left( 1-\frac{v^{2}}{c^{2}} \right)}} \) the equivalent relation in electricity for electric charge is:
q = q0
\[q=\frac{q_{0}}{\sqrt{\left( 1-\frac{v^{2}}{c^{2}} \right)}}\]
\[ q_{0}=\frac{q}{\sqrt{\left( 1-\frac{v^{2}}{c^{2}} \right)}}\]
\[ q=\frac{q_{0}v}{c}\]

Solution:

The relativistic formula for mass,

 

m=m01v2c2m = \frac{m_0}{\sqrt{1 - \frac{v^2}{c^2}}}

 

accounts for how mass increases with velocity. This behavior arises because mass is a form of energy, and energy is affected by motion under relativity.

However, electric charge (

qq

) is invariant under relativistic mechanics. Charge does not depend on the velocity of the particle. It remains constant in all inertial reference frames, which is a fundamental principle in physics.

Thus, the equivalent relation for electric charge is simply:

 

q=q0q = q_0

 

This reflects the fact that charge does not vary with velocity, unlike mass.

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