Ratio of electric and magnetic field due of moving point charge if its speed is $4.5 \times 10^{5} \;m / s$
AIIMS 2019, Diffcult
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The ratio of electric and magnetic field is given by,
$\frac{E}{B}=\frac{\frac{K Q}{r^{2}}}{\frac{\mu_{0}}{4 \pi}\left(\frac{q V}{r^{2}}\right)}$
$=\frac{K Q 4 \pi}{\mu_{0} q V}$
$=\frac{1}{4 \pi \varepsilon_{0}}\left(\frac{4 \pi}{\mu_{0}}\right)$
$=\frac{C^{2}}{V}$ $....(I)$
Substitute $3 \times 10^{8}$ for $C$ and $4.5 \times 10^{5}$ for $V$ in equation $( I )$.
$\frac{E}{B}=\frac{3 \times 10^{8}}{4.5 \times 10^{5}}$
$=2 \times 10^{11}$
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