Home
Class 12
PHYSICS
A uniform electric field E is created be...

A uniform electric field E is created between two parallel
., charged plates as shown in figure . An electron
. enters the field symmetrically between the plataes with a
. speed `v_0. The length of each plate is l. Find the angle of
. deviation of the path of the electron as it comes out
. of the field.
θ = tan − 1 E l m v 2 0 θ = tan − 1 ( e E l m v 2 0 ) θ = tan − 1 ( e E l m v 0 ) θ = tan − 1 ( e E m v 2 0 )

Text Solution

Verified by Experts

The correct Answer is:
The electron deviates by an angle
`theta="tan"^(-1) (eEl)/(mv_(0)^(2))` (from x-axis)`=45^(º)`
Promotional Banner

Topper's Solved these Questions

  • ELECTROSTATICS

    RESONANCE ENGLISH|Exercise Exercise-1 Section (C)|14 Videos
  • ELECTROSTATICS

    RESONANCE ENGLISH|Exercise Exercise-1 Section (D)|5 Videos
  • ELECTROSTATICS

    RESONANCE ENGLISH|Exercise Exercise-1 Section (A)|9 Videos
  • ELECTROMAGNETIC INDUCTION

    RESONANCE ENGLISH|Exercise A.l.P|19 Videos
  • EXPERIMENTAL PHYSICS

    RESONANCE ENGLISH|Exercise PART -II|10 Videos

Similar Questions

Explore conceptually related problems

A uniform electric field of strength e exists in a region. An electron (charge -e, mass m) enters a point A with velocity V hat(j) . It moves through the electric field & exits at point B. Then :

A parallel plate condenser has a unifrom electric field E (V//m) in the space between the plates. If the distance between the plates is d(m) and area of each plate is A(m^(2)) the energy (joule) stored in the condenser is

A parallel plate condenser has a unifrom electric field E (V//m) in the space between the plates. If the distance between the plates is d(m) and area of each plate is A(m^(2)) the energy (joule) stored in the condenser is

A particle of mass m and charge -q enters the region between the two charged plates initially moving along x-axis with speed v_(x) as shown in figure. The length of plate is L and a uniform electric field E is maintained between the plates. The vertical deflection of the particle at the far edge of the plate is

An electron projectes with velocity vecv=v_(0)hati in the electric field vecE=E_(0)hatj . There the path followed by the electron E_(0) .

In uniform electric field, E = 10 NC^(-1) as shown in figure, find (i) V_(A)-V_(B) (ii) V_(B)-V_(C) .

If uniform electric field vec E = E_ 0 hati + 2 E _ 0 hatj , where E _ 0 is a constant exists in a region of space and at (0,0) the electric potential V is zero, then the potential at ( x _ 0 , 2x _ 0 ) will be -

Five identical conducting plates each of face area .A. are arranged as shown in figure. If a potential. difference of .V._(0) is created between plate A and C then the charge on plate .E. will be [Given C = (epsilon_0A)/d ]

A particle of charge q and mass m starts moving from the origin under the action of an electric field vec E = E_0 hat i and vec B = B_0 hat i with a velocity vec v = v_0 hat j . The speed of the particle will become 2v_0 after a time.

A particle of charge q and mass m starts moving from the origin under the action of an electric field vec E = E_0 hat i and vec B = B_0 hat i with a velocity vec v = v_0 hat j . The speed of the particle will become 2v_0 after a time.