Figure shows a schematic view of an electrostatic analyzer. It can sort out charged particles by speed and charge to mass ratio. Spacecraft use such analyzers to characterize charged particles in interplanetary space. Two curved metal plates establish an electric field given by E = E_{0} \left(\right. \frac{b}{r} \left.\right) where E0 and b are positive constants with unit of electric field and length. The field points toward the centre of curvature and r is distance from centre. There is no influence of gravity. Proton (charge +e mass 'm') enters along y-axis and exits along x-axis while moving along a circular path.

Figure shows a schematic view of an electrostatic analyzer. It can sort out charged particles by speed and charge to mass ratio. Spacecraft use such analyzers to characterize charged particles in interplanetary space. Two curved metal plates establish an electric field given by $E = E_{0} (\frac{b}{r})$ where E₀ and b are positive constants with unit of electric field and length. The field points toward the centre of curvature and r is distance from centre. There is no influence of gravity. Proton (charge +e mass 'm') enters along y-axis and exits along x-axis while moving along a circular path.

Linked Question 1

Speed with which proton is to be projected is 'v' and centripetal acceleration of electron is 'a_c' is given by respectively. Mark the correct statement

$v = \sqrt{\frac{2 {eE}_{0} b}{m}}; a_{c} = \frac{e}{2 m} E_{0} (\frac{b}{r})$

$v = \sqrt{\frac{{eE}_{0} b}{m}}; a_{c} = \frac{2 e}{m} E_{0} (\frac{b}{r})$

$v = \sqrt{\frac{{eE}_{0} b}{2 m}}; a_{c} = \frac{2 e}{m} E_{0} (\frac{b}{r})$

$v = \sqrt{\frac{{eE}_{0} b}{m}}; a_{c} = \frac{e}{m} E_{0} (\frac{b}{r})$

$eE = \frac{{mv}^{2}}{r} \Rightarrow v = \sqrt{\frac{{eE}_{0} b}{m}}$
$a_{c} = \frac{eE}{m} = \frac{e}{m} E_{0} (\frac{b}{r})$

Linked Question 2

Mark the correct option.

If proton enters closer to the inner surface it will require smaller speed to follow circular trajectory.

It does not matter where the protons enter the device it requires same speed to follow circular trajectory.

If E₀ is made larger then in order to maintain same trajectory initial speed has to be decreased.

A deuteron (charge +e, mass 2m) will require greater speed as compared to proton to follow circular trajectory.

v is independent of r.

Linked Question 3

Mark the incorrect option :

If $v = \sqrt{\frac{2 {eE}_{0} b}{m}}$ proton may strike inner surface of analyzer.

If $v = \sqrt{\frac{2 {eE}_{0} b}{m}}$ proton may strike outer surface of analyzer.

Work done by electric field on proton is zero.

If an electron is released with zero initial velocity from inner surface of analyzer it will strike outer surface with velocity $v = \sqrt{\frac{2 {eE}_{0} b}{m_{e}} ℓn (\frac{b}{a})}$ , where me is mass of electron.

If $v = \sqrt{\frac{2 {eE}_{0} b}{m}}$ , centrifugal force > electric force (centripetal force) in the reference frame of proton.
$\frac{1}{2} m_{e} v^{2} = {eE}_{0} b \int_{a}^{b} \frac{dr}{r}$ $\Rightarrow v = \sqrt{\frac{2 {eE}_{0} b}{m_{e}} ℓn (\frac{b}{a})}$