Q. 644.7( 3 Votes )
An electric field of magnitude 1000 NC–1 is produced between two parallel plates having a separation of 2.0 cm as shown in figure.
(a) What is the potential difference between the plate?
(b) With what minimum speed should an electron be projected from the lower plate in the direction of the field so that it may reach the upper plate?
(c) Suppose the electron is projected from the lower plate with the speed calculated in part (b). The direction of projection makes an angle of 60° with the field. Find the maximum height reached by the electron.
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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(a) In a quark model of elementary particles, a neutron is made of one up quarks [charge (2/3) e] and two down quarks [charges – (1/3) e]. Assume that they have a triangle configuration with side length of the order of 10–15 m. Calculate electrostatic potential energy of neutron and compare it with its mass 939 MeV.
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