Q. 643.8( 4 Votes )
An electric field
Magnitude of the electric field : E = 1000 NC–1
Separation between the plates: r = 2 cm = 0.02 m
Angle made by the projection with the field : θ = 60°
The Potential difference is:
V = -E.r
Here, E is the electric field and r is the separation between the plates.
V = - 1000 × 0.2
∴ V = -200 = |-200| = 200 V
Hence, the potential difference between the plates is of 200 V
Charge on an electron: e = -1.6× 10-19 C
We know that
Where F is the electric force, E is the electric field, m is the mass of the body and a is the acceleration of the body.
Here, q=e of electron. And m = 9.7× 10-31 kg mass of the electron.
Using one of the equations of motion we get,
Here, v is the final velocity of the electron= 0
Here v is zero as it we have to calculate u when it just reaches the upper plate, u is the initial velocity of the electron, s is the distance between the plates: s=r and a is the acceleration of the electrons
Hence, with a minimum speed of 2.64× 106 m/s2 should the electron be projected from the lower plate in the direction of the field for it to reach the upper plate.
As direction of projection makes an angle 60° with the electric field.
Initial velocity is resolved into its cos component which is along the direction of the electric field as shown in the figure above.
Hence, u’=ucos(60° ), u’ is the resolved initial velocity.
Using the same equation of motion used in (b) we get,
Where h is the maximum height reached by the electron.
Hence, the maximum height reached by the electron is 4.7× 10-3 m.
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