Two electrons are moving with equal speed of 5 × 10 ^ 5 ms^ - 1...

Given, speed of electrons, $v = 5 \times {10}^{5}ms^{- 1}$

Electrostatic force between the electrons,

${F}_{e} = k \cdot \dfrac{e2}{r2}... \left(i\right)$

Magnetic field produced due to moving electron with velocity $v$,

$B = \dfrac{\mu0}{4\pi} \cdot \dfrac{ev}{r2}...\left(ii\right)$

Magnetic force between the electrons,

${F}_{m} = Bev = \dfrac{\mu0}{4\pi} \cdot \dfrac{ev}{r2} \cdot ev$ & (from Eq. (ii))

${F}_{m} = \dfrac{\mu0}{4\pi} \cdot \dfrac{e^{2}v^{2}}{r2}$

$∴$ From Eqs. (i) and (iii), we have

$\dfrac{Fe}{Fm} = \dfrac{ke^{2} / r^{2}}{{μ}_{0}e^{2}v^{2}4\pi \cdot r^{2}} = \dfrac{k \cdot 4\pi}{{μ}_{0}v^{2}}$

$= \dfrac{9 \times {10}^{9} \times 4 \times 3. 14}{4\pi \times {10}^{- 7}\left(5 \times {10}^{5}\right)^{2}} = 3. 6 \times {10}^{5}$