An electron enters uniform electric field maintained by parallel plates and of value EVm-1 with a velocity v text ms-1, the plates are separated by a distance d metre, then acceleration of the electron in the field is:

Question

An electron enters uniform electric field maintained by parallel plates and of value EVm-1 with a velocity v text ms-1, the plates are separated by a distance d metre, then acceleration of the electron in the field is: (A)  (Ee/m)  (B)  (-Ee/m)  (C)  (Ed/md)  (D)  Ee(d/m)

Tardigrade Admin · Accepted Answer

F=eE Therefore, acceleration in electron due to force F a=(F/m)=(eE/m)

Q. An electron enters uniform electric field maintained by parallel plates and of value $E V m^{- 1}$ with a velocity $v m s^{- 1},$ the plates are separated by a distance $d$ metre, then acceleration of the electron in the field is:

$\frac{E e}{m}$

$\frac{- E e}{m}$

$\frac{E d}{m d}$

$E e \frac{d}{m}$

$\frac{E m}{e d}$

$F = e E$
Therefore, acceleration in electron due to force F
$a = \frac{F}{m} = \frac{e E}{m}$

Q. An electron enters uniform electric field maintained by parallel plates and of value EVm−1 with a velocity vms−1, the plates are separated by a distance d metre, then acceleration of the electron in the field is:

mEe​

m−Ee​

mdEd​

Eemd​

edEm​

F=eE Therefore, acceleration in electron due to force F a=mF​=meE​