An electron is released from the bottom plate A as shown in the figure `(E = 10^(4) N//C)`. The velocity of the electron when it reaches plate B will be nearly equal to

The maximum velocity of the electron emitted from the photosensitive plate is determined by the

A block of A is released from rest from the top of wedge block of height h shown in figure-4.55. If velocity of block when it reaches the bottom of inclive is v_(0) , find the time os sliding.

A uniform magnetic field (B) is applied between the plates of a capacitor in the outward direction as shown in the figure. Surface charge density on the capacitor plates is sigma . An electron is fired in between the plates, parallel to them, as shown in the figure. The electron is found to move straight in-between the plates of capacitor. Neglect the gravity. If L is the length of the plates then caleulate the time taken by the electron to cross the plates.

In the uniform electric field of E = 1 xx 10^(4) NC^(-1) an electron is accelerated from rest . The velocity of the electron when it has travelled a distance of 2 xx 10^(-2) m is nearly … ms^(-1) ( e/m of electron = 1.8 xx 10^(11) C kg^(-1) )

Two long parallel plates A and B are separated by a distance of 4 cm with an electric field of 45 . 5 Vm^(-1) between the plates normally from plate A to plate B , as shown in the figure. An electron is projected from plate A with velocity v at an angle of 30^(@) with the surface of plate A . The maximum value of v so that the electron does not hit plate B is (Assume gravity free space, charge of electron = 1.6 xx 10^(-19) C and mass of electron = 9 . 1 xx 10^(-31) kg )

Two charged metal plates in vacuum are 10cm apart. A uniform electric field of intensity (45//16)xx10^(3)NC^(-1) is applied between the plates. An electron is released between the plates from rest at a point just outside the negative plate. (a) Calculate how long (t) will the electron take to reach the other plate? (b) At what velocity (v) will it be electron take to hits teh other plate?

Two charged metal plates in vacuum are 10cm apart. A uniform electric field of intensity (45//16)xx10^(3)NC^(-1) is applied between the plates. An electron is released between the plates from rest at a point just outside the negative plate. (a) Calculate how long (t) will the electron take to reach the other plate? (b) At what velocity (v) will it be electron take to hits teh other plate?