The electron gun in an old TV picture tube accelerates electrons between two parallel plates 1.6 cm apart with a 20 kV potential difference between them. The electrons enter through a small hole in the negative plate, accelerate, then exit through a small hole in the positive plate. Assume that the holes are small enough not to affect the electric field or potential. a)What is the electric field strength between the plates?

b)With what speed does an electron exit the electron gun if its entry speed is close to zero? Note: the exit speed is so fast that we really need to use the theory of relativity to compute an accurate value. Your answer to part B is in the right range but a little too big.

(a) E = 1.25×10⁶ V/m

(b) v = 83.8×10⁶ m/s

Explanation:

Please see attachment below.

qV = electric potential energy of the electron. This is the energy converted into the kinetic energy of the motion of the electron through the plates. Kinetic energy is 1/2×mv². Where m = mass of electron = 9.11×10-³¹ kg.

q = 1.6×10-¹⁹ C.

So qV = 1/2mv² so rearranging and substituting the given quantities give the speed of the electrons.

here we know that ball will drop down by total distance 3.2 m

so here we know that

now using kinematics

now in the same time it will cover horizontal distance of 0.75 m

so speed of the ball will be

wire in brush may conduct electricity from cell to drill and vice versa electrc shock possible