Question

An electron at rest has a charge of $1.6\times {{10}^{-19}}\,C$. It starts moving with a velocity of $v=\dfrac{c}{2}$, where $c$ is the speed of light ,then the new charge value on it is-
A. $1.6\times {{10}^{-19}}\,Coulomb$
B. $1.6\times {{10}^{-19}}{{\sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}}}^{{}}}Coulomb$
C. $1.6\times {{10}^{-19}}{{\sqrt{{{\left( \dfrac{1}{2} \right)}^{2}}-1}}^{{}}}Coulomb$
D. ${{\dfrac {1.6\times {{10}^{-19}}}{\sqrt{1-{{\left( \dfrac{1}{2} \right)}^{2}}}}}^{{}}}Coulomb$

Answer 1

Hint: Relativistic mass which is a special theory of relativity to be used here. All the physical properties like force and momentum are assumed to be based on rest mass ${{m}_{o}}$ of the body. But when a body moves when a body moves with certain speed, we find that mass changes by a factor of $\gamma $making the new relativistic mass $m=\dfrac{{{m}_{o}}}{\gamma }$ where \[\gamma =\sqrt[{}]{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}\] ,$c$ is speed of light.

Complete answer:
Considering the relativistic mass formula, we can see that mass is a function of speed but not the charge of a particle. Theoretically if the speed of a particle approaches the speed of light mass becomes infinite. The variation due to speed is observed in the mass but overall charge of the particle is conserved.it is independent from the frame of reference. Charge is not considered in relativity and hence is not variant.as the mass increases so the charge density may differ due to volume difference but the overall charge remains constant.

So, the charge on the electron will remain $1.6\times {{10}^{-19}}\,Coulomb$ which is option A.

Note:
Photon’s rest mass is considered as$0$and no relativistic mass. The quantities measured from an observer view are not confined to mass.
Two major observations made by the observer are ageing rate and body’s length in the direction of motion of the body, both of which reduces by a factor of $\gamma $ which is $\gamma =\sqrt[{}]{1-\dfrac{{{v}^{2}}}{{{c}^{2}}}}$,which is the concept of relativistic mass.