Cathode rays of velocity 10 ^ 6 ^ -1 describe an approximate...

Cathode rays are composed of electrons, when they move in electric field a force

$F = eE$...(i)

Acts on them, this provides the necessary centripetal force to the particles

$F = \dfrac{mv^{2}}{r}$...(ii)

From Eqs. (i) and (ii), we get

$eE = \dfrac{mv^{2}}{r}$

$⇒ r = \dfrac{mv^{2}}{eE} = \dfrac{m{\left({10}^{6}\right)}^{2}}{e ( 300 )}$...(iii)

When velocity is doubled same circular path is followed, hence radius is same

$r = \dfrac{m{\left(2 \times {10}^{6}\right)}^{2}}{eE}$...(iv)

Equating Eqs. (iii) and (iv), we get

$m \times \dfrac{{\left({10}^{6}\right)}^{2}}{300e} = \dfrac{m \times {\left(2 \times {10}^{6}\right)}^{2}}{eE}$

$⇒ E = 300 \times 4 = 1200$ V $cm^{- 1}$